Solving Linear Equations
| Site: | Young Education |
| Cours: | Linear Equations and Inequalities |
| Livre: | Solving Linear Equations |
| Imprimé par: | Guest user |
| Date: | vendredi 25 septembre 2026, 02:38 |
1. Review of Algebraic Equations
Learning outcomes
- I can identify variables, constants, coefficients, and terms in an equation.
- I can explain the meaning of equality in an equation.
- I can solve simple one-step and two-step equations.
- I can verify whether a value is a solution to an equation.
- I can apply inverse operations to isolate variables.
2. Equations with Variables on Both Sides
Learning outcomes
- I can identify equations that contain variables on both sides.
- I can simplify equations by combining like terms.
- I can move variable terms to one side of an equation.
- I can solve equations with variables on both sides accurately.
- I can check and verify solutions.
Introduction
In earlier lessons, you learned to solve equations where the variable appeared on only one side of the equal sign. However, many real-world problems produce equations in which the variable appears on both sides. These equations require one additional step: moving all the variable terms to one side before solving.
Although these equations may look more complicated, they follow exactly the same principle as simpler equations—you must keep the equation balanced by performing the same operation on both sides. Once all the variables are collected on one side, the equation becomes familiar and can be solved using inverse operations.
What Are Equations with Variables on Both Sides?
An equation with variables on both sides has one or more variable terms on each side of the equal sign.
Examples:
Unlike one-step or two-step equations, you must first collect all the variable terms together before solving.
Step 1 – Simplify Each Side
If either side contains like terms, combine them before solving.
Like terms have exactly the same variable raised to the same power.
For example:
Example
Simplify:
Combine the like terms:
This makes the equation easier to solve.
Visualising Like Terms
Like-term tiles help show why only terms with the same variable can be combined.
Step 2 – Move the Variable Terms
Choose one side to contain all the variables.
Most students find it easiest to move the smaller variable term.
Worked Example 1
Solve:
Step 1
Subtract x from both sides.
Step 2
Subtract 4.
Step 3
Divide by 2.
Step 3 – Solve the Remaining Equation
Once the variables are on one side, solve using the same methods learned previously.
Worked Example 2
Solve:
Subtract
2x:
Subtract 7:
Divide by 3:
Example with Variables on Both Sides and Negative Numbers
Negative numbers require careful attention to signs.
Worked Example 3
Solve:
Subtract 2x:
Add 8:
Divide by 4:
Always perform the same operation on both sides to keep the equation balanced.
Example Requiring Like Terms First
Sometimes you must simplify before moving variables.
Worked Example 4
Solve:
Step 1
Combine like terms.
Step 2
Subtract x.
Step 3
Subtract 5.
Step 4
Divide by 4.
Checking Your Solution
Always substitute your answer back into the original equation.
Example
Original equation:
Solution:
Check the left side:
Check the right side:
Both sides equal 13, so the solution is correct.
Special Cases
Not every equation has exactly one solution.
Identity
Example:
Subtract 4x:
This statement is always true.
Answer: Every value of x is a solution.
No Solution
Example:
Subtract 3x:
This statement is never true.
Answer: No solution exists.
Summary Table
| Final Result | Meaning |
|---|---|
| One solution | |
| Infinitely many solutions (identity) | |
| No solution |
Common Mistakes
Students often make these errors.
❌ Moving a term without changing its sign.
✔ Remember that adding or subtracting a term changes its position only because you perform the operation on both sides.
❌ Forgetting to combine like terms first.
✔ Simplify each side whenever possible before solving.
❌ Making arithmetic mistakes with negative numbers.
✔ Work carefully and write each step clearly.
❌ Forgetting to check the final answer.
✔ Substitute the value into the original equation, not a simplified version.
Real-World Connection
Equations with variables on both sides appear in many practical situations. For example, a mobile phone company may charge a monthly fee plus a cost per gigabyte of data. Comparing two different pricing plans involves setting the total costs equal and solving for the amount of data used. Engineers, scientists, economists, and computer programmers frequently solve equations of this type when comparing competing models or determining when two quantities become equal.
Did You Know?
Many computer algebra systems, including those used by engineers and scientists, solve equations by following the same logical steps you learn in school: simplify each side, collect like terms, isolate the variable, and verify the solution. Although computers perform these steps much faster, they are applying the same mathematical principles.
Key Terms
- Equation — a mathematical statement showing that two expressions are equal.
- Variable — a symbol representing an unknown value.
- Like terms — terms with the same variable raised to the same power.
- Simplify — combine like terms or reduce an expression to its simplest form.
- Inverse operations — operations that undo each other.
- Identity — an equation that is true for every value of the variable.
- No solution — an equation with no value that satisfies it.
- Solution — a value that makes an equation true.
Key Takeaways
- Equations with variables on both sides require collecting all variable terms on one side before solving.
- Always simplify each side by combining like terms first.
- Use inverse operations to move variables and constants while keeping the equation balanced.
- Some equations have one solution, some have infinitely many solutions, and some have no solution.
- Always verify your answer by substituting it into the original equation.
3. Equations with Fractions and Decimals
Learning outcomes
- I can solve equations containing decimal coefficients.
- I can solve equations containing fractional coefficients.
- I can eliminate fractions using appropriate multiplication techniques.
- I can simplify equations before solving.
- I can verify solutions involving fractions and decimals.
4. Literal Equations and Formula Rearrangement
Learning outcomes
- I can identify variables and constants within formulas.
- I can rearrange formulas to isolate a specified variable.
- I can apply inverse operations when manipulating formulas.
- I can solve scientific and geometric formulas for different variables.
- I can verify the correctness of a rearranged formula.
What Is a Literal Equation?
A literal equation is an equation containing two or more variables.
Many formulas used in mathematics, science, engineering, and finance are literal equations.
Examples include:
d = vt
F = ma
A = πr²
V = IR
I = Prt
Instead of solving for a numerical answer, we often rearrange the formula so that a particular variable is isolated.
For example:
d = vt
can be rearranged to:
v = d/t
or:
t = d/v
The relationship has not changed. We have simply written the formula in a different form.
Variables and Constants
A variable is a quantity that can change.
For example, in:
d = vt
the letters:
d, v, and t
represent variables.
They might represent:
- d = distance
- v = velocity
- t = time
Each quantity can take different values in different situations.
A constant is a value that remains fixed.
For example, in:
C = 2πr
the number:
2
is a constant.
The value:
π
is also a mathematical constant, approximately:
3.14159
The variables are:
C and r
Coefficients in Formulas
A coefficient is a number or expression multiplying a variable.
For example:
y = 3x
The coefficient of x is:
3
In:
A = 1/2 bh
the coefficient of bh is:
1/2
In:
C = 2πr
the coefficient multiplying r is:
2π
Understanding what is multiplying or dividing the target variable is essential when rearranging formulas.
What Does It Mean to Rearrange a Formula?
To rearrange a formula means to rewrite it so that a chosen variable is alone on one side.
This is also called:
making a variable the subject of the formula.
For example:
F = ma
already has F as the subject.
If we want m as the subject:
m = F/a
If we want a as the subject:
a = F/m
The formula still represents the same physical relationship.
The Balance Principle Still Applies
Formula rearrangement uses exactly the same algebra rules as ordinary equation solving.
Whatever operation is performed on one side must also be performed on the other side.
For example:
A = bh
Suppose we want to isolate h.
h is multiplied by b.
Use the inverse operation:
divide by b
So:
A/b = bh/b
Therefore:
h = A/b
The other letters are treated just like known numbers while we isolate the target variable.
Inverse Operations
Formula rearrangement depends heavily on inverse operations.
| Operation | Inverse Operation |
|---|---|
| Addition | Subtraction |
| Subtraction | Addition |
| Multiplication | Division |
| Division | Multiplication |
| Squaring | Square root |
| Square root | Squaring |
The goal is to undo the operations surrounding the variable.
Example 1: Addition
Rearrange:
y = x + 7
Solve for x.
Subtract 7 from both sides:
y − 7 = x
Therefore:
x = y − 7
Example 2: Subtraction
Rearrange:
A = B − C
Solve for C.
Add C to both sides:
A + C = B
Subtract A:
C = B − A
Example 3: Multiplication
Rearrange:
P = IV
Solve for I.
I is multiplied by V.
Divide both sides by V:
P/V = I
Therefore:
I = P/V
Example 4: Division
Rearrange:
v = d/t
Solve for d.
Multiply both sides by t:
vt = d
Therefore:
d = vt
This is one reason formula triangles can be useful for simple multiplication-and-division relationships.
Rearranging Distance, Speed and Time
The basic motion formula is:
d = vt
where:
- d = distance
- v = velocity or speed
- t = time
To solve for velocity:
d = vt
Divide by t:
v = d/t
To solve for time:
d = vt
Divide by v:
t = d/v
So the three equivalent forms are:
d = vt
v = d/t
t = d/v
Formula Triangles
Formula triangles can help with simple relationships involving multiplication and division.
For example:
The density formula is:
ρ = m/V
where:
- ρ = density
- m = mass
- V = volume
From the same relationship:
m = ρV
and:
V = m/ρ
Formula triangles are useful shortcuts, but algebraic rearrangement is more powerful because it works with much more complicated formulas.
Rearranging the Density Formula
Start with:
ρ = m/V
Solve for m.
Multiply both sides by V:
ρV = m
Therefore:
m = ρV
Now solve for V.
Start again:
ρ = m/V
Multiply by V:
ρV = m
Divide by ρ:
V = m/ρ
One formula can therefore answer three different questions.
Ohm's Law
A common electrical formula is:
V = IR
where:
- V = voltage
- I = current
- R = resistance
To solve for current:
V = IR
Divide by R:
I = V/R
To solve for resistance:
V = IR
Divide by I:
R = V/I
Rearranging Formulas with More Than One Operation
Some formulas require several inverse operations.
Consider:
y = 3x + 5
Solve for x.
First subtract 5:
y − 5 = 3x
Then divide by 3:
x = (y − 5)/3
Notice the order.
We undo the addition first, then the multiplication.
This is often the reverse of the order in which the operations act on x.
Work from the Outside In
Consider:
P = 4x + 12
To isolate x:
First remove the addition:
P − 12 = 4x
Then remove multiplication by 4:
x = (P − 12)/4
A useful strategy is:
identify what is being done to the target variable and undo those operations in reverse order.
Example with Brackets
Rearrange:
y = 5(x + 2)
Solve for x.
First divide both sides by 5:
y/5 = x + 2
Then subtract 2:
x = y/5 − 2
Equivalent form:
x = (y − 10)/5
Both are correct.
Example with the Variable in a Denominator
Rearrange:
v = d/t
Solve for t.
Because t is in the denominator, first multiply both sides by t:
vt = d
Then divide by v:
t = d/v
When the target variable appears in a denominator, it is often helpful to remove the fraction first.
Geometry Example: Area of a Triangle
The area of a triangle is:
A = 1/2 bh
where:
- A = area
- b = base
- h = perpendicular height
Suppose we want h as the subject.
Start:
A = 1/2 bh
Multiply both sides by 2:
2A = bh
Divide by b:
h = 2A/b
So:
h = 2A/b
Rearranging the Triangle Formula for Base
Start with:
A = 1/2 bh
Multiply by 2:
2A = bh
Divide by h:
b = 2A/h
So the formula can be written as:
A = 1/2 bh
b = 2A/h
h = 2A/b
Geometry Example: Area of a Circle
The area of a circle is:
A = πr²
Suppose we want to solve for r.
Start:
A = πr²
Divide both sides by π:
A/π = r²
Now take the square root:
r = √(A/π)
This example introduces another inverse pair:
square ↔ square root
Why Do We Need a Square Root?
In:
A = πr²
the variable r is squared.
To isolate r, we must undo the square.
The inverse operation of:
r²
is:
√
Therefore:
r = √(A/π)
For a geometric radius, we use the positive root because radius represents a non-negative length.
Pythagorean Theorem
The Pythagorean theorem is:
a² + b² = c²
Usually c represents the hypotenuse.
If we want c:
a² + b² = c²
Take the square root:
c = √(a² + b²)
Rearranging the Pythagorean Theorem for a
Start with:
a² + b² = c²
Subtract b²:
a² = c² − b²
Take the square root:
a = √(c² − b²)
Similarly:
b = √(c² − a²)
Scientific Example: Newton's Second Law
Newton's second law is:
F = ma
where:
- F = resultant force
- m = mass
- a = acceleration
Solve for mass:
m = F/a
Solve for acceleration:
a = F/m
This is a good example of how algebra allows one scientific formula to calculate several different quantities.
Scientific Example: Kinetic Energy
Kinetic energy is given by:
Eₖ = 1/2 mv²
Suppose we want mass.
Start:
Eₖ = 1/2 mv²
Multiply by 2:
2Eₖ = mv²
Divide by v²:
m = 2Eₖ/v²
Rearranging Kinetic Energy for Velocity
Start:
Eₖ = 1/2 mv²
Multiply by 2:
2Eₖ = mv²
Divide by m:
2Eₖ/m = v²
Take the square root:
v = √(2Eₖ/m)
This requires three stages:
- remove 1/2
- remove m
- undo the square
Temperature Conversion Formula
A common relationship between Celsius and Fahrenheit temperatures is:
F = 9/5 C + 32
Suppose we want C as the subject.
Start:
F = 9/5 C + 32
Subtract 32:
F − 32 = 9/5 C
Multiply by 5/9:
C = 5/9(F − 32)
This gives the familiar Celsius conversion formula.
Why the Brackets Matter
Notice:
C = 5/9(F − 32)
The entire quantity:
F − 32
must be multiplied by 5/9.
It would be incorrect to write:
C = 5/9F − 32
because that represents a different relationship.
Brackets help preserve the correct order of operations.
Finance Example: Simple Interest
The simple interest formula is:
I = Prt
where:
- I = interest
- P = principal
- r = interest rate
- t = time
The formula is commonly used to calculate simple interest.
Rearranging Simple Interest for Rate
Start:
I = Prt
We want r.
Divide both sides by Pt:
I/(Pt) = r
Therefore:
r = I/(Pt)
Rearranging Simple Interest for Time
Start:
I = Prt
Divide both sides by Pr:
t = I/(Pr)
For principal:
P = I/(rt)
One formula can therefore be rearranged depending on which quantity is unknown.
When the Target Variable Appears More Than Once
Some formulas are more challenging because the target variable appears in several terms.
Consider:
ax + bx = c
We want x.
We cannot simply divide by a or b because x appears twice.
First factor x:
x(a + b) = c
Now divide by:
a + b
Therefore:
x = c/(a + b)
Factoring is an important strategy in literal equations.
Another Factoring Example
Rearrange:
P = xy + xz
Solve for x.
Factor x:
P = x(y + z)
Divide by:
y + z
Therefore:
x = P/(y + z)
Variable on Both Sides
Consider:
y = ax + bx
This is straightforward because x can be factored:
y = x(a + b)
Therefore:
x = y/(a + b)
But sometimes the variable appears on both sides.
Example:
ax + b = cx + d
Solve for x.
Move the x terms together:
ax − cx = d − b
Factor:
x(a − c) = d − b
Divide:
x = (d − b)/(a − c)
provided:
a − c ≠ 0
Formula Rearrangement and Restrictions
Whenever we divide by an expression, that expression must not equal zero.
For example:
x = c/(a + b)
requires:
a + b ≠ 0
because division by zero is undefined.
At an introductory level, these restrictions may not always be emphasized, but they become increasingly important in advanced algebra.
Fractional Formulas
Consider:
A = (x + y)/z
Solve for x.
Multiply both sides by z:
Az = x + y
Subtract y:
x = Az − y
If solving for z:
Start:
A = (x + y)/z
Multiply by z:
Az = x + y
Divide by A:
z = (x + y)/A
Example with Several Terms
Rearrange:
v = u + at
Solve for a.
Subtract u:
v − u = at
Divide by t:
a = (v − u)/t
This formula appears frequently in motion problems.
Rearranging for Time
Start:
v = u + at
Subtract u:
v − u = at
Divide by a:
t = (v − u)/a
The same formula can therefore be used to calculate:
- final velocity
- initial velocity
- acceleration
- time
depending on which variable is unknown.
Rearranging for Initial Velocity
Start:
v = u + at
Subtract at:
v − at = u
Therefore:
u = v − at
This requires only one inverse operation.
A Systematic Rearrangement Method
A reliable approach is:
- Identify the variable you need to isolate.
- Identify all operations acting on that variable.
- Simplify or factor if necessary.
- Apply inverse operations.
- Perform the same operation on both sides.
- Continue until the target variable is alone.
- Rewrite with the target variable on the left if desired.
- Verify the rearranged formula.
Example Using the Method
Rearrange:
A = 3x + 2y
Solve for x.
Step 1: Target variable
x
Step 2: Operations involving x
x is:
- multiplied by 3
- then 2y is added
Step 3: Undo addition
A − 2y = 3x
Step 4: Undo multiplication
x = (A − 2y)/3
Verifying a Rearranged Formula
How can we know that a rearranged formula is correct?
One method is to rearrange it back to the original form.
Suppose:
Original:
F = ma
Rearranged:
a = F/m
Multiply both sides by m:
ma = F
Rewrite:
F = ma
We have recovered the original formula.
Therefore the rearrangement is consistent.
Verification by Substitution
Another method is to choose numerical values.
Suppose:
F = ma
Let:
m = 4
and:
a = 3
Then:
F = 4 × 3 = 12
Now use the rearranged formula:
a = F/m
Substitute:
a = 12/4
a = 3
The same result is obtained.
This provides a useful check.
Verify the Triangle Formula
Original:
A = 1/2 bh
Let:
b = 10
h = 6
Then:
A = 1/2(10)(6)
A = 30
Our rearranged formula was:
h = 2A/b
Substitute:
h = 2(30)/10
h = 6
The original value is recovered.
Therefore the rearrangement works.
Dimensional and Unit Checks
Scientific formulas can sometimes be checked using units.
Consider:
v = d/t
Units are:
m/s = m/s
Now rearrange:
d = vt
Units:
m = (m/s)(s)
The seconds cancel:
m = m
This supports the rearrangement.
Unit Check for F = ma
Original:
F = ma
Units:
N = kg × m/s²
Now rearrange:
m = F/a
Units:
kg = N ÷ (m/s²)
Since:
1 N = 1 kg·m/s²
the units simplify to:
kg
A unit check can sometimes reveal algebraic errors.
Common Mistake: Moving Terms Without Explaining the Operation
Students often say:
"Move the 5 to the other side."
For:
y = 3x + 5
it is better to think:
subtract 5 from both sides
giving:
y − 5 = 3x
Terms do not magically change sides.
Their signs change because an inverse operation was performed.
Common Mistake: Dividing Only One Term
Suppose:
A − B = 3x
Then:
x = (A − B)/3
Both A and B are divided by 3.
It would be incorrect to write:
x = A − B/3
unless brackets are handled appropriately.
Common Mistake: Forgetting Brackets
Correct:
x = (y − 5)/3
Incorrect:
x = y − 5/3
The two expressions are not equivalent.
Brackets show that the entire numerator is divided by 3.
Common Mistake: Forgetting the Square Root
From:
r² = A/π
it is incorrect to conclude:
r = A/π
You must undo the square:
r = √(A/π)
Common Mistake: Square Rooting Only One Term
From:
c² = a² + b²
correct:
c = √(a² + b²)
not:
c = a + b
In general:
√(a² + b²) ≠ a + b
Common Mistake: Changing the Formula Instead of Rearranging It
A rearranged formula must describe exactly the same relationship as the original.
For example:
Original:
F = ma
Correct:
m = F/a
Incorrect:
m = Fa
The second formula represents a different mathematical relationship.
Why Formula Rearrangement Matters
Formula rearrangement allows us to use one relationship in many different ways.
For example:
V = IR
can calculate voltage if current and resistance are known.
But in a different experiment, voltage and resistance may be known and current may be unknown.
Rather than memorizing three separate formulas, we can rearrange one:
V = IR
I = V/R
R = V/I
This is much more flexible.
Application in Science
Science is full of formulas that need rearranging.
Examples include:
v = d/t
F = ma
ρ = m/V
V = IR
Eₖ = 1/2 mv²
p = F/A
Q = mcΔT
Being able to manipulate formulas reduces the number of separate equations that need to be memorized.
Application in Geometry
Geometry also depends heavily on formula rearrangement.
Examples include:
A = bh
A = 1/2 bh
A = πr²
C = 2πr
V = lwh
V = 1/3 πr²h
The variable you need depends on the information given in the problem.
Example: Circumference of a Circle
Start with:
C = 2πr
Solve for r.
Divide by 2π:
r = C/(2π)
If:
C = 31.4 cm
then:
r = 31.4/(2π)
approximately:
r = 5.0 cm
Example: Volume of a Rectangular Prism
The volume formula is:
V = lwh
Solve for h.
Divide both sides by lw:
h = V/(lw)
Similarly:
l = V/(wh)
and:
w = V/(lh)
Challenge Example: Cone Volume
The volume of a cone is:
V = 1/3 πr²h
Solve for h.
Multiply both sides by 3:
3V = πr²h
Divide by πr²:
h = 3V/(πr²)
Now solve for r.
Start:
3V = πr²h
Divide by πh:
3V/(πh) = r²
Take the square root:
r = √(3V/(πh))
Challenge Example: Rearranging with the Variable on Both Sides
Solve for x:
ax + b = cx + d
Subtract cx:
ax − cx + b = d
Subtract b:
ax − cx = d − b
Factor x:
x(a − c) = d − b
Divide:
x = (d − b)/(a − c)
This combines several algebra skills:
- collecting variable terms
- factoring
- dividing
- using brackets correctly
Did You Know?
Scientists and engineers often rearrange formulas before entering values into a calculator.
For example, if:
Eₖ = 1/2 mv²
and velocity is required, it is usually clearer to rearrange first:
v = √(2Eₖ/m)
and only then substitute the numbers.
This reduces repeated calculations and makes the mathematical structure easier to check.
A Useful Rearrangement Checklist
Before finishing, ask:
Is the requested variable completely alone?
Did I perform the same operation on both sides?
Did I use inverse operations correctly?
Do I need brackets?
Did I factor if the target variable appeared more than once?
Did I undo powers using roots?
Could any denominator become zero?
Can I rearrange the result back to the original formula?
If the answer to these questions is yes, the rearrangement is likely correct.
Key Terms
Literal equation – An equation containing two or more variables.
Formula – An equation describing a relationship between quantities.
Variable – A quantity whose value can change.
Constant – A fixed value.
Coefficient – A number or expression multiplying a variable.
Subject of a formula – The variable isolated on one side of a formula.
Rearrange – Rewrite a formula so that a different variable becomes the subject.
Inverse operation – An operation that reverses another operation.
Isolate – Get a variable alone on one side of an equation.
Factor – Rewrite an expression as a product of factors.
Verify – Check that a rearranged formula is equivalent to the original.
Key Takeaways
- A literal equation contains several variables.
- Variables represent quantities that may change.
- Constants have fixed values.
- Rearranging a formula means making a chosen variable the subject.
- Formula rearrangement follows the same balance rules as ordinary equations.
- Use inverse operations to isolate the target variable.
- Addition is undone by subtraction.
- Multiplication is undone by division.
- Squaring is undone using a square root.
- If the target variable appears more than once, you may need to factor it.
- Brackets are important when an entire expression is divided or multiplied.
- Scientific formulas such as F = ma, V = IR, and ρ = m/V can be rearranged for different quantities.
- Geometric formulas such as A = πr² and A = 1/2 bh can also be rearranged.
- The same formula can often answer several different types of questions.
- You can verify a rearrangement by transforming it back into the original equation.
- Numerical substitution and unit checks can also help verify the result.
- A reliable strategy is:
identify the target → simplify if needed → undo operations → isolate the variable → verify.
5. Real-World Equation Modeling
Learning outcomes
- I can translate verbal descriptions into algebraic equations.
- I can identify variables and unknown quantities in real situations.
- I can solve equations that model real-world problems.
- I can interpret solutions within the context of a problem.
- I can evaluate whether a solution is reasonable.
What Is Equation Modeling?
A mathematical model is a representation of a real situation using mathematics.
A real-world problem might describe:
- the cost of a taxi ride
- the number of tickets sold
- the distance travelled
- the dimensions of a shape
- a phone bill
- a savings goal
- an age relationship
- a measurement
- a salary plus commission
The words tell us how quantities are related.
Our job is to translate those relationships into an equation.
Once the equation is written, we can use algebra to solve it.
A useful overall process is:
real situation → identify unknown → write equation → solve → interpret → check
A reliable word-problem strategy begins by identifying what is unknown, assigning a variable, translating the relationships into an equation, solving, and then checking the result in context.
Why Is Modeling Important?
Algebra is useful because it allows us to represent situations that contain an unknown quantity.
For example:
A streaming service charges a monthly fee of $8 plus $2 per movie.
Your total bill is $22.
How many movies did you rent?
We could guess and check.
But algebra gives us a systematic approach.
Let:
m = number of movies
The cost of the movies is:
2m
The fixed monthly fee is:
8
Total bill:
22
So the equation is:
2m + 8 = 22
Subtract 8:
2m = 14
Divide by 2:
m = 7
Interpretation:
7 movies were rented.
This is equation modeling.
Start by Identifying the Unknown
Before writing an equation, determine:
What am I trying to find?
This unknown quantity should be represented by a variable.
For example:
A concert ticket costs $18. You spent $126. How many tickets did you buy?
The unknown is:
number of tickets
We could write:
Let t = number of tickets
Then:
18t = 126
Solve:
t = 7
The variable has a specific meaning in the problem.
It is not simply an abstract letter.
Define Your Variable Clearly
Good modeling begins with a clear variable statement.
Instead of writing only:
x = ?
write:
Let x = number of notebooks purchased
or:
Let d = distance travelled in kilometres
or:
Let w = width of the garden in metres
This helps prevent confusion later.
A variable should represent a quantity, not a unit or vague idea.
Known Quantities vs Unknown Quantities
Every problem contains information that is already known and something that must be found.
Consider:
A gym charges a joining fee of $25 and $15 each month. Your total cost was $130.
Known quantities:
- joining fee = $25
- monthly fee = $15
- total cost = $130
Unknown:
- number of months
Let:
m = number of months
Then:
25 + 15m = 130
Recognising these quantities is often more important than immediately looking for arithmetic keywords.
Translating Words into Mathematics
Certain words often suggest mathematical operations.
| Words or Phrase | Possible Operation |
|---|---|
| sum, total, added to, increased by | Addition |
| difference, decreased by, less than | Subtraction |
| product, times, twice, triple | Multiplication |
| quotient, divided by, per | Division |
| is, equals, results in | Equals |
These words are helpful clues, but they should not replace understanding the situation.
Addition Phrases
Examples:
A number plus 7
becomes:
x + 7
Eight more than a number
becomes:
x + 8
A number increased by 12
becomes:
x + 12
The total of a number and 5
becomes:
x + 5
Subtraction Phrases
Examples:
A number minus 4
becomes:
x − 4
A number decreased by 9
becomes:
x − 9
But be careful with:
5 less than a number
This means:
x − 5
not:
5 − x
Order matters in subtraction.
"Less Than" Can Reverse the Order
Consider:
3 less than 10
We calculate:
10 − 3
Therefore:
3 less than x
means:
x − 3
Similarly:
7 less than twice a number
means:
2x − 7
This is a common source of errors in equation modeling.
Multiplication Phrases
Examples:
three times a number
3x
twice a number
2x
half a number
x/2
the product of 5 and a number
5x
Remember:
twice x = 2x
not:
x²
Division Phrases
Examples:
a number divided by 5
x/5
the quotient of a number and 4
x/4
half of a number
x/2
a cost of $24 shared equally among x people
24/x
The order of division matters.
From an Expression to an Equation
An expression does not contain an equals sign.
Example:
3x + 5
An equation states that two expressions are equal.
Example:
3x + 5 = 20
Real-world modeling often requires us to determine what quantity represents each side of the equation.
Example: Translating a Verbal Statement
"The sum of twice a number and 9 is 31."
Let:
x = the number
Twice the number:
2x
Add 9:
2x + 9
"is 31" means:
= 31
Equation:
2x + 9 = 31
Solve:
2x = 22
x = 11
Not Every Problem Should Be Translated Word by Word
Keywords are helpful, but modeling is really about relationships.
For example:
"A taxi charges $4 to begin a ride and $2.50 for every kilometre travelled. The total fare is $19."
We should think:
total cost = starting fee + distance charge
Let:
d = distance in kilometres
Then:
19 = 4 + 2.50d
or equivalently:
4 + 2.50d = 19
Understanding the structure is more reliable than hunting for individual keywords.
A Useful Modeling Framework
For many linear situations, the model has the form:
Total = fixed amount + rate × quantity
or:
y = mx + b
Here:
- b represents a fixed starting amount
- m represents a rate per unit
- x represents the number of units
- y represents the total
This pattern appears in many real-life problems.
Examples include:
- taxi fares
- phone plans
- hourly wages
- delivery fees
- equipment rentals
- memberships
Example: Taxi Fare
A taxi charges:
$5 starting fee
plus:
$3 per kilometre
The total fare is:
$29
Let:
k = kilometres travelled
Model:
5 + 3k = 29
Subtract 5:
3k = 24
Divide by 3:
k = 8
Interpretation:
The taxi travelled 8 km.
Does the Answer Make Sense?
We should check:
Starting fee:
$5
Distance charge:
8 × $3 = $24
Total:
$5 + $24 = $29
Yes.
The solution satisfies both the equation and the real-world situation.
Example: Phone Plan
A phone plan costs:
$20 per month
plus:
$0.10 per text message above the included limit
One month the bill was:
$32.50
Let:
t = number of extra text messages
Equation:
20 + 0.10t = 32.50
Subtract 20:
0.10t = 12.50
Divide:
t = 125
Interpretation:
125 extra messages were sent.
Check:
$20 + $0.10(125)
= $20 + $12.50
= $32.50
Reasonable.
Example: Hourly Pay
A worker receives a base payment of $45 plus $18 per hour.
The total pay for a job was:
$153
Let:
h = number of hours worked
Equation:
45 + 18h = 153
Subtract 45:
18h = 108
Divide by 18:
h = 6
Interpretation:
The worker worked for 6 hours.
Example: Savings
You already have:
$120
You save:
$25 each week
You want:
$320
Let:
w = number of weeks
Model:
120 + 25w = 320
Subtract 120:
25w = 200
Divide:
w = 8
It will take:
8 weeks
to reach the goal.
The Meaning of the Variable Matters
Suppose the algebra gave:
w = 8
That answer alone is incomplete.
What does 8 mean?
It might mean:
- $8
- 8 km
- 8 people
- 8 weeks
- 8 tickets
In the savings example:
w = 8 weeks
Always interpret the solution in the context of the problem.
Distance, Rate and Time Problems
A common real-world relationship is:
distance = rate × time
or:
d = rt
This formula can be used to create and solve linear equations.
Example: Distance Travelled
A cyclist travels at:
18 km/h
for:
t hours
and covers:
72 km
Equation:
18t = 72
Divide by 18:
t = 4
Interpretation:
The cyclist travelled for 4 hours.
Check:
18 km/h × 4 h = 72 km
Example: Journey with a Fixed Distance
A driver has already travelled:
45 km
and continues at:
70 km/h
The total journey is:
255 km
How much longer must the driver travel?
Let:
t = remaining travel time in hours
Distance still being modelled:
70t
Equation:
45 + 70t = 255
Subtract 45:
70t = 210
Divide:
t = 3
The driver must travel:
3 more hours.
Geometry as a Real-World Model
Equations can also model physical dimensions.
Suppose a rectangular garden has:
width = x metres
length = x + 4 metres
perimeter = 32 metres
The perimeter formula is:
P = 2L + 2W
Substitute:
32 = 2(x + 4) + 2x
Expand:
32 = 2x + 8 + 2x
Combine:
32 = 4x + 8
Subtract 8:
24 = 4x
Therefore:
x = 6
Width:
6 m
Length:
10 m
Check the Geometry Answer
Perimeter:
2(10) + 2(6)
= 20 + 12
= 32
So the solution is correct.
Also, both dimensions are positive, which is necessary for a real rectangle.
Modeling Consecutive Numbers
Real-world modeling can also involve numbers that have a special relationship.
Consecutive integers differ by 1.
If the first integer is:
x
then the next is:
x + 1
and the next:
x + 2
Suppose two consecutive integers have a sum of 41.
Equation:
x + (x + 1) = 41
Simplify:
2x + 1 = 41
2x = 40
x = 20
Therefore the numbers are:
20 and 21
Modeling Ages
Suppose Maya is 6 years older than Ben.
Together their ages total 34.
Let:
b = Ben's age
Then Maya's age is:
b + 6
Equation:
b + (b + 6) = 34
Simplify:
2b + 6 = 34
2b = 28
b = 14
Maya:
14 + 6 = 20
Answer:
Ben is 14 and Maya is 20.
Why Parentheses Are Useful
In the previous example:
b + (b + 6) = 34
The parentheses show that:
b + 6
represents Maya's entire age.
Parentheses become especially important when a quantity is multiplied.
Example:
A family buys 3 tickets, each costing:
x + 4 dollars
Total:
3(x + 4)
This is different from:
3x + 4
Example: Ticket Sales
Adult tickets cost:
$12
Student tickets cost:
$7
Suppose a group buys:
3 adult tickets
and some student tickets.
The total cost is:
$71
Let:
s = number of student tickets
Adult cost:
3(12) = 36
Student cost:
7s
Equation:
36 + 7s = 71
Subtract 36:
7s = 35
Therefore:
s = 5
The group bought:
5 student tickets.
Integer Restrictions Matter
Suppose you solve a ticket problem and obtain:
s = 5.4
Mathematically, the equation may have been solved correctly.
But in the real-world context:
5.4 tickets does not make sense
because tickets are normally counted as whole objects.
This tells us to reconsider:
- whether the equation was written correctly
- whether the data were copied correctly
- whether rounding is appropriate
- whether the problem expects an integer solution
Context matters.
Example: Buying Boxes
A school needs at least 100 markers.
Markers are sold in boxes of 12.
If a calculation says:
8.33 boxes
you cannot normally purchase 0.33 of a sealed box.
So the practical answer would be:
9 boxes
This shows that interpreting a model may require more than simply reporting a calculator result.
Continuous Quantities Are Different
Some quantities can reasonably contain decimals.
Examples include:
- distance
- mass
- time
- volume
- money
- temperature
For example:
t = 2.5 hours
is perfectly reasonable.
Likewise:
d = 7.25 km
can make sense.
So whether a decimal answer is acceptable depends on the quantity being modeled.
Example: Restaurant Bill
Three friends split a bill equally.
The bill includes a fixed service charge of:
$9
The total bill is:
$72
Suppose each person's food cost before the service charge is the same.
Let:
x = food cost per person
Equation:
3x + 9 = 72
Subtract 9:
3x = 63
Divide by 3:
x = 21
Each person's food cost was:
$21
Example: Temperature Relationship
Suppose a laboratory sample begins at:
18°C
and warms at:
4°C per minute
After some time, its temperature reaches:
46°C
Let:
t = time in minutes
Model:
18 + 4t = 46
Subtract 18:
4t = 28
t = 7
The sample heated for:
7 minutes.
Model the Relationship, Not Just the Numbers
A common mistake is to see numbers and immediately perform arithmetic.
Consider:
A parking garage charges $6 to enter and $2 for every hour parked. Your bill is $18.
Numbers:
6, 2, 18
Blind arithmetic might lead to:
18 ÷ 6
or:
18 ÷ 2
Neither fully represents the situation.
The correct relationship is:
starting fee + hourly charge = total
So:
6 + 2h = 18
This gives:
h = 6
Understanding the story comes first.
Fixed Costs and Variable Costs
Many real-world linear equations contain two kinds of costs.
Fixed cost
A cost that does not depend on how many units are used.
Examples:
- entrance fee
- subscription fee
- equipment rental fee
- delivery charge
Variable cost
A cost that changes with the number of units.
Examples:
- cost per kilometre
- cost per hour
- cost per item
- cost per gigabyte
This often creates:
Total cost = fixed cost + variable rate × quantity
Example: Bicycle Rental
A bicycle rental shop charges:
$12 fixed fee
plus:
$6 per hour
Your bill is:
$42
Let:
h = hours rented
Equation:
12 + 6h = 42
Subtract 12:
6h = 30
Divide:
h = 5
The bicycle was rented for:
5 hours.
Reverse Modeling
Sometimes you are given a situation and must decide whether an equation correctly represents it.
Suppose:
"A music service charges $10 per month plus $0.50 per downloaded song."
Let:
s = number of downloaded songs
Which equation gives total cost C?
Correct:
C = 10 + 0.50s
Why?
- $10 is fixed
- $0.50 depends on the number of songs
Incorrect:
C = 10s + 0.50
because that would charge $10 per song.
Units Can Help You Build an Equation
Units are powerful clues.
Suppose:
$4 per notebook × n notebooks
The notebooks cancel conceptually, leaving:
dollars
So:
4n
represents a cost.
Likewise:
60 km/h × t hours
gives:
60t km
which represents distance.
If the units of your expression do not match the quantity being modeled, something may be wrong.
Example: Unit Reasoning
A car travels:
80 km/h
for:
t hours
Distance:
80 km/h × t h
The hours cancel:
80t km
Therefore:
d = 80t
Unit analysis supports the equation.
Interpreting Negative Solutions
Suppose a problem asks:
"How many tickets were sold?"
and your solution is:
x = −12
That is not physically reasonable.
You cannot sell negative tickets in the ordinary interpretation.
A negative result may indicate:
- the model was written incorrectly
- the given numbers are inconsistent
- a sign was reversed
- the variable was defined differently than expected
Not every algebraic solution is meaningful in every context.
Interpreting Zero
Zero can sometimes be completely reasonable.
For example:
"How many additional kilometres must you travel?"
A solution:
d = 0
could mean:
you have already arrived.
Context determines whether zero makes sense.
Evaluating Reasonableness with Estimation
You do not always need an exact calculation to spot an unreasonable result.
Suppose:
A notebook costs about $4.
You spent about $40.
You should expect roughly:
10 notebooks
If your algebra gives:
x = 100
something is probably wrong.
Estimation provides a useful reality check.
Example: Estimating First
A concert charges:
$24 per ticket
plus a one-time booking fee of:
$5
The total is:
$101
Before solving:
Four tickets cost about:
4 × $24 = $96
plus $5:
$101
So we already expect:
4 tickets
Now solve:
24t + 5 = 101
24t = 96
t = 4
The exact solution agrees with our estimate.
Checking by Substitution
Once an equation is solved, substitute the solution back into the model.
Suppose:
3x + 8 = 35
Solution:
x = 9
Check:
3(9) + 8
= 27 + 8
= 35
Mathematically correct.
Then check the context.
If x represented:
number of books
then:
9 books
is reasonable.
This gives two kinds of checking:
algebraic check
and:
context check
Algebraically Correct Does Not Always Mean Contextually Correct
Suppose:
A boat can carry 6 people per row.
An equation gives:
r = 4.5 rows
The algebra may be correct numerically.
But a real boat cannot normally have half of a seating row added for one trip.
The final interpretation may require:
5 rows
depending on the problem.
Mathematical models represent reality, but interpretation still requires judgment.
Multi-Step Modeling Example
A museum charges:
$14 admission
plus:
$3 for each special exhibit
A visitor spends:
$26
How many special exhibits did the visitor see?
Let:
e = number of special exhibits
Equation:
14 + 3e = 26
Subtract 14:
3e = 12
Divide by 3:
e = 4
Interpretation:
The visitor saw 4 special exhibits.
Check:
14 + 3(4) = 26
Reasonable because 4 exhibits is a whole, non-negative number.
More Challenging Example: Perimeter
A rectangular field is 7 m longer than it is wide.
Its perimeter is 54 m.
Let:
w = width
Then:
length = w + 7
Using:
P = 2L + 2W
we get:
54 = 2(w + 7) + 2w
Expand:
54 = 2w + 14 + 2w
Combine:
54 = 4w + 14
Subtract 14:
40 = 4w
w = 10
Length:
17
Therefore:
width = 10 m
length = 17 m
Check:
2(17) + 2(10) = 54
More Challenging Example: Two Related Quantities
There are 38 students in two classes.
Class A has 6 more students than Class B.
Let:
x = number of students in Class B
Then Class A has:
x + 6
Equation:
x + (x + 6) = 38
Simplify:
2x + 6 = 38
2x = 32
x = 16
So:
Class B = 16
Class A = 22
Check:
16 + 22 = 38
and:
22 − 16 = 6
Both conditions are satisfied.
Modeling Requires All Conditions to Be Satisfied
The previous example gives an important idea.
A solution must satisfy every condition in the problem.
It is not enough that:
16 + 22 = 38
We also need:
22 is 6 more than 16
A good contextual check tests all important relationships.
A Structured Problem-Solving Method
A strong approach to real-world equations is:
1. Read
Understand the situation.
2. Identify
Determine what is known and unknown.
3. Define
Choose a variable and state what it represents.
4. Model
Translate the relationships into an equation.
5. Solve
Use algebra to isolate the variable.
6. Interpret
State what the numerical solution means.
7. Evaluate
Check the equation and decide whether the answer is reasonable.
This closely matches standard algebraic problem-solving methods.
Visualizing the Modeling Process
A verbal statement contains mathematical structure.
For example:
"Five less than twice a number is 17."
Break it apart:
twice a number:
2x
five less than this:
2x − 5
is 17:
2x − 5 = 17
Now solve:
2x = 22
x = 11
Common Mistake: Choosing the Wrong Variable
Suppose:
"A jacket is $15 cheaper than a pair of shoes. Together they cost $105."
If:
x = cost of the shoes
then jacket cost:
x − 15
Equation:
x + (x − 15) = 105
But if:
x = cost of the jacket
then shoes cost:
x + 15
Equation:
x + (x + 15) = 105
Both approaches can work.
The important thing is to define the variable clearly and remain consistent.
Common Mistake: Ignoring a Fixed Amount
Problem:
A delivery service charges $8 plus $3 per package.
For 5 packages:
Incorrect:
3(5) = 15
Correct:
8 + 3(5) = 23
The fixed charge must be included.
Common Mistake: Multiplying the Wrong Quantity
"A gym charges $20 membership plus $10 per class."
Incorrect:
20x + 10
Correct:
20 + 10x
Only the class charge depends on the number of classes.
Common Mistake: Giving Only a Number
Suppose:
x = 12
If x represents months, the final answer should say:
It will take 12 months.
If x represents kilograms:
The mass is 12 kg.
Context is part of the answer.
Common Mistake: Trusting the Calculator Automatically
A calculator may correctly solve the arithmetic while the original model is wrong.
If you write the wrong equation, the calculator will faithfully solve:
the wrong equation.
That is why the modeling step is often the most important part.
Real-World Models Have Assumptions
Mathematical models simplify reality.
For example:
C = 5 + 2d
for a taxi fare assumes:
- the starting fee is exactly $5
- the rate remains $2 per kilometre
- there are no traffic charges
- there are no tolls
- there are no special surcharges
A model can be useful without including every possible detail.
Understanding its assumptions helps us know when the model is appropriate.
When a Linear Model Works
Linear equations work well when a quantity changes at a constant rate.
Examples:
- $4 per item
- 60 km each hour
- 3 cm growth each week
- $15 saved each month
These situations follow a pattern such as:
y = mx + b
where the rate of change remains constant.
When a Linear Model May Not Work
Not every real-world situation is linear.
Examples might include:
- compound interest
- population growth
- falling objects over long periods
- curved trajectories
- areas involving squared dimensions
For example:
The area of a circle is:
A = πr²
Because r is squared, the relationship is not linear.
Choosing the correct type of model is part of mathematical thinking.
Did You Know?
When engineers, economists, scientists, and businesses create mathematical models, they rarely begin by solving equations.
They first decide:
- which quantities matter
- which quantities can change
- which relationships are important
- which assumptions are reasonable
The equation comes after the situation has been understood.
That is why real-world equation modeling is more than an algebra skill. It is also a reasoning skill.
Model-Building Example
Suppose a company rents equipment.
The cost is:
- $50 setup fee
- $18 per day
We can create a general model.
Let:
d = number of days
Let:
C = total cost
Then:
C = 50 + 18d
This model can now answer many questions.
For 3 days:
C = 50 + 18(3)
C = 104
If total cost is $176:
176 = 50 + 18d
126 = 18d
d = 7
One equation can model an entire family of situations.
From a Table to an Equation
Suppose a delivery company has this pricing pattern:
| Packages | Cost |
|---|---|
| 1 | $11 |
| 2 | $14 |
| 3 | $17 |
| 4 | $20 |
The cost increases by:
$3 per package
So the rate is:
3
If one package costs $11, then the fixed starting cost must be:
$8
because:
8 + 3(1) = 11
Model:
C = 8 + 3p
This shows that real-world equations can be developed from data as well as written descriptions.
A Graph Can Represent the Same Model
For:
C = 8 + 3p
the equation, table, graph, and verbal description all represent the same relationship.
The 8 represents the starting cost.
The 3 represents the cost added per package.
This is an important connection between linear equations and linear functions.
Evaluating Whether a Model Makes Sense
After solving, ask:
- Is the value positive when it should be?
- Should the answer be a whole number?
- Are the units correct?
- Is the size of the answer reasonable?
- Does substitution reproduce the original information?
- Does the answer satisfy all conditions?
- Did I include every fixed amount and rate?
These questions turn calculation into mathematical reasoning.
Worked Example: Complete Modeling Process
A kayak rental costs $18 plus $7 per hour. A customer pays $53.
Step 1: Identify the unknown
Number of hours.
Step 2: Define the variable
Let h = number of hours rented.
Step 3: Identify the relationship
Total cost:
fixed fee + hourly cost
Step 4: Write the equation
18 + 7h = 53
Step 5: Solve
Subtract 18:
7h = 35
Divide by 7:
h = 5
Step 6: Interpret
The kayak was rented for 5 hours.
Step 7: Check
18 + 7(5)
= 18 + 35
= 53
The result is positive, realistic, and satisfies the original problem.
Challenge Example
A rectangular garden has a length that is 3 m more than twice its width.
Its perimeter is:
42 m
Let:
w = width
Then:
length = 2w + 3
Perimeter formula:
P = 2L + 2W
Substitute:
42 = 2(2w + 3) + 2w
Expand:
42 = 4w + 6 + 2w
Combine:
42 = 6w + 6
Subtract 6:
36 = 6w
Therefore:
w = 6
Length:
2(6) + 3 = 15
Check:
2(15) + 2(6)
= 30 + 12
= 42
Answer:
The garden is 6 m wide and 15 m long.
Key Terms
Model – A mathematical representation of a real situation.
Variable – A symbol representing a quantity that can change or is unknown.
Unknown – A quantity whose value must be determined.
Equation – A statement showing that two mathematical expressions are equal.
Expression – A mathematical combination of numbers, variables, and operations without an equals sign.
Fixed amount – A quantity that remains constant regardless of the number of units.
Rate – A quantity measured per unit, such as dollars per hour or kilometres per hour.
Interpret – Explain what a mathematical result means in its original context.
Reasonable solution – A solution that is mathematically correct and sensible in the real situation.
Constraint – A condition that limits possible values, such as requiring a number of people to be a whole number.
Assumption – A condition treated as true when building a model.
Key Takeaways
- Real-world equation modeling translates everyday situations into mathematics.
- Begin by identifying what quantity is unknown.
- Clearly define a variable and state what it represents.
- Separate known quantities from unknown quantities.
- Look for relationships between quantities rather than relying only on keywords.
- Addition phrases include total, increased by, sum, and more than.
- Subtraction phrases include difference, decreased by, and less than.
- Multiplication phrases include times, product, twice, and triple.
- Division phrases include quotient, divided by, and per.
- Be careful with order in expressions such as 5 less than x = x − 5.
- Many linear real-world models have the structure:
total = fixed amount + rate × quantity
- Use algebra to solve the model.
- Always interpret the numerical solution using the appropriate units and context.
- Whole-number quantities such as people, tickets, or boxes may require special interpretation.
- Decimal answers are often appropriate for time, distance, mass, and other continuous quantities.
- Substitute your answer back into the equation to check the mathematics.
- Also check whether the result makes sense in reality.
- Estimation can help identify unreasonable solutions.
- Units can help verify that an equation has been modeled correctly.
- A mathematically correct solution may still be meaningless if it violates the real-world conditions.
- Good modeling follows:
understand → define → translate → solve → interpret → evaluate.