Solving Systems by Substitution

A system of equations is a set of two or more equations containing the same variables. A solution to the system must make all equations in the system true at the same time.

For example:

y = x + 2

2x + y = 8

The solution is:

(2, 4)

because x = 2 and y = 4 satisfy both equations.

The substitution method solves a system by replacing one variable with an equivalent expression from the other equation.

The main idea is:

Solve → Substitute → Solve → Substitute Back → Verify

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What Does Substitution Mean?

To substitute means to replace something with an equivalent value or expression.

Suppose:

y = x + 3

Then anywhere we see y, we can replace it with:

x + 3

For example, if another equation is:

2x + y = 13

substitute x + 3 for y:

2x + (x + 3) = 13

Now the equation contains only one variable.

That is the key purpose of substitution.

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Why Substitution Works

Consider:

y = x + 3

This equation tells us that y and x + 3 have the same value.

Therefore, replacing y with x + 3 does not change the mathematical relationship.

In:

2x + y = 13

we can write:

2x + (x + 3) = 13

The two equations have now been combined into one equation involving only x.

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When Is Substitution Appropriate?

Substitution can solve almost any linear system, but it is especially efficient when:

• one variable is already isolated
• one equation is written as y = ...
• one equation is written as x = ...
• a variable has coefficient 1
• a variable has coefficient −1
• isolating a variable requires very little work

For example:

y = 3x − 4

2x + y = 11

is ideal for substitution because y is already isolated.

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Recognizing a Good Substitution System

Consider:

x = 2y + 1

3x − y = 13

The first equation already tells us exactly what x equals.

Therefore, substitute:

2y + 1

for x in the second equation.

Substitution is usually more efficient here than rearranging both equations into another form.

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A Less Convenient Example

Consider:

4x + 7y = 23

5x − 6y = 8

Neither variable is already isolated, and solving for one variable immediately creates fractions.

Substitution still works, but another method such as elimination may be more efficient.

Choosing an efficient method is part of solving systems successfully.

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The Substitution Method

A reliable procedure is:

Isolate → Substitute → Solve → Substitute Back → Verify

Isolate

Choose one equation and solve for one variable if necessary.

Substitute

Replace that variable in the other equation with its equivalent expression.

Solve

Solve the resulting one-variable equation.

Substitute Back

Use the value you found to determine the other variable.

Verify

Substitute both values into both original equations.

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Example: Variable Already Isolated

Solve:

y = x + 2

2x + y = 8

Since y is already isolated:

y = x + 2

Substitute x + 2 for y:

2x + (x + 2) = 8

Combine like terms:

3x + 2 = 8

Subtract 2:

3x = 6

Divide by 3:

x = 2

Now substitute x = 2 into:

y = x + 2

y = 2 + 2

y = 4

Therefore:

(x, y) = (2, 4)

or simply:

(2, 4)

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Why the Answer Is an Ordered Pair

A system involving x and y usually has a solution written as:

(x, y)

So:

x = 2

y = 4

becomes:

(2, 4)

The order matters.

(2, 4) means:

x = 2, y = 4

It does not mean x = 4 and y = 2.

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Verify the Solution

For:

y = x + 2

2x + y = 8

test:

x = 2, y = 4

First equation:

4 = 2 + 2

4 = 4 ✓

Second equation:

2(2) + 4 = 8

8 = 8 ✓

Therefore:

(2, 4)

is the solution to the system.

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The Graphical Meaning

Each linear equation represents a straight line.

The solution to the system is the point where the two lines intersect.

For:

y = x + 2

and:

2x + y = 8

the lines intersect at:

(2, 4)

Substitution finds this intersection algebraically, without needing to estimate it from a graph.

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Example with x Already Isolated

Solve:

x = y + 3

2x + y = 12

Substitute y + 3 for x:

2(y + 3) + y = 12

Distribute:

2y + 6 + y = 12

Combine like terms:

3y + 6 = 12

Subtract 6:

3y = 6

Therefore:

y = 2

Now substitute back:

x = y + 3

x = 2 + 3

x = 5

Solution:

(5, 2)

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Parentheses Are Important

Suppose:

y = 2x − 5

and:

3x − y = 9

Substitute the entire expression 2x − 5 for y:

3x − (2x − 5) = 9

Parentheses are essential because the minus sign applies to the whole expression.

Distribute the negative:

3x − 2x + 5 = 9

Therefore:

x + 5 = 9

x = 4

Substitute back:

y = 2(4) − 5

y = 3

Solution:

(4, 3)

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A Common Substitution Error

For:

3x − y = 9

and:

y = 2x − 5

incorrect substitution would be:

3x − 2x − 5 = 9

The correct substitution is:

3x − (2x − 5) = 9

which becomes:

3x − 2x + 5 = 9

Whenever an expression replaces a variable, putting the substituted expression in parentheses is a very good habit.

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When a Variable Is Not Yet Isolated

Consider:

x + y = 9

2x + 3y = 23

Neither variable is written alone, but the first equation is easy to rearrange.

From:

x + y = 9

solve for x:

x = 9 − y

Now substitute into the second equation:

2(9 − y) + 3y = 23

Expand:

18 − 2y + 3y = 23

Combine:

18 + y = 23

Therefore:

y = 5

Substitute back:

x = 9 − 5

x = 4

Solution:

(4, 5)

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Which Variable Should You Isolate?

In:

x + y = 9

we could solve for either variable.

For x:

x = 9 − y

For y:

y = 9 − x

Both are equally simple.

Choose whichever makes the next substitution easiest.

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Look for a Coefficient of 1

Consider:

3x + y = 14

2x − 4y = −8

The first equation contains y with coefficient 1.

So solve for y:

y = 14 − 3x

This is easier than solving the second equation for x or y because those choices would introduce fractions.

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Complete Example

Solve:

3x + y = 14

2x − 4y = −8

From the first equation:

y = 14 − 3x

Substitute into the second:

2x − 4(14 − 3x) = −8

Distribute:

2x − 56 + 12x = −8

Combine:

14x − 56 = −8

Add 56:

14x = 48

Therefore:

x = 24/7

Now substitute back:

y = 14 − 3(24/7)

Write 14 as 98/7:

y = 98/7 − 72/7

y = 26/7

Solution:

(24/7, 26/7)

A system does not need to have whole-number solutions.

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Negative Coefficient of 1

A variable with coefficient −1 can also be easy to isolate.

Consider:

2x − y = 7

Solve for y.

Subtract 2x:

−y = 7 − 2x

Multiply by −1:

y = 2x − 7

Now this expression can be substituted into the other equation.

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Example with Negative Values

Solve:

y = −2x + 1

3x + y = −7

Substitute:

3x + (−2x + 1) = −7

Combine:

x + 1 = −7

Therefore:

x = −8

Substitute back:

y = −2(−8) + 1

y = 16 + 1

y = 17

Solution:

(−8, 17)

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Checking Negative Solutions

First equation:

17 = −2(−8) + 1

17 = 17 ✓

Second equation:

3(−8) + 17 = −7

−24 + 17 = −7

−7 = −7 ✓

Therefore:

(−8, 17)

is correct.

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Substitution with Fractions

Solve:

y = x/2 + 1

x + y = 7

Substitute:

x + (x/2 + 1) = 7

Subtract 1:

x + x/2 = 6

Write x as:

2x/2

Then:

2x/2 + x/2 = 6

3x/2 = 6

Multiply by 2:

3x = 12

Therefore:

x = 4

Substitute back:

y = 4/2 + 1

y = 3

Solution:

(4, 3)

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Substitution with Decimals

Solve:

y = 0.5x + 2

x + y = 8

Substitute:

x + (0.5x + 2) = 8

Combine:

1.5x + 2 = 8

Subtract 2:

1.5x = 6

Therefore:

x = 4

Substitute back:

y = 0.5(4) + 2

y = 4

Solution:

(4, 4)

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Both Equations Written as y = ...

Substitution is particularly convenient when both equations are already solved for the same variable.

Consider:

y = 2x + 1

y = −x + 7

Since both expressions equal y:

2x + 1 = −x + 7

Add x:

3x + 1 = 7

Subtract 1:

3x = 6

Therefore:

x = 2

Substitute:

y = 2(2) + 1

y = 5

Solution:

(2, 5)

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Why We Can Set the Expressions Equal

If:

y = 2x + 1

and:

y = −x + 7

then both:

2x + 1

and:

−x + 7

represent the same value of y at the solution.

Therefore:

2x + 1 = −x + 7

This is substitution even though the replacement step can look like simply setting two expressions equal.

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A More Complex Example

Solve:

y = 3x − 8

2x + 5y = 9

Substitute:

2x + 5(3x − 8) = 9

Distribute:

2x + 15x − 40 = 9

Combine:

17x − 40 = 9

Add 40:

17x = 49

Therefore:

x = 49/17

Substitute back:

y = 3(49/17) − 8

y = 147/17 − 136/17

y = 11/17

Solution:

(49/17, 11/17)

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Substitution Can Produce No Solution

Consider:

y = 2x + 3

2y = 4x + 10

Substitute:

2(2x + 3) = 4x + 10

Expand:

4x + 6 = 4x + 10

Subtract 4x:

6 = 10

This is false.

Therefore:

No solution

The equations represent parallel lines.

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Why a False Statement Means No Solution

The substitution process eliminated the variables and produced:

6 = 10

There is no value of x or y that can make this statement true.

Therefore there is no ordered pair satisfying both equations.

The lines never intersect.

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Substitution Can Produce Infinitely Many Solutions

Consider:

y = 2x + 3

2y = 4x + 6

Substitute:

2(2x + 3) = 4x + 6

Expand:

4x + 6 = 4x + 6

This is always true.

Therefore:

Infinitely many solutions

The two equations represent the same line.

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Why a True Statement Means Infinitely Many Solutions

If substitution produces:

4x + 6 = 4x + 6

then every point satisfying one equation also satisfies the other.

The equations describe the same line.

Therefore the system has infinitely many solutions.

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Three Possible Outcomes

A system of two linear equations can have:

Recognizing these outcomes is an important part of substitution.

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Verifying a Solution

Suppose your solution is:

(3, 5)

and the equations are:

x + y = 8

2x − y = 1

Check the first:

3 + 5 = 8 ✓

Check the second:

2(3) − 5 = 1

6 − 5 = 1 ✓

Because both equations are satisfied:

(3, 5)

is a solution to the system.

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Why Check Both Equations?

Checking only one equation is not enough.

A point may lie on one line but not the other.

For example:

x + y = 8

The point:

(2, 6)

satisfies this equation.

But if the second equation is:

2x − y = 1

then:

2(2) − 6 = −2

not 1.

Therefore:

(2, 6)

is not a solution to the system.

A system solution must satisfy every equation.

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Finding an Error by Verification

Suppose a student claims that:

(4, 2)

solves:

y = x − 2

2x + y = 9

Check the first:

2 = 4 − 2 ✓

Check the second:

2(4) + 2 = 10

But the equation requires 9.

Therefore:

(4, 2)

is not the solution.

Verification helps catch arithmetic and substitution errors.

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Writing Clear Substitution Work

Good algebraic communication makes each replacement visible.

For example:

y = 3x + 2

2x + y = 17

Write:

2x + (3x + 2) = 17

rather than jumping immediately to:

5x + 2 = 17

The first line clearly shows where the substitution occurred.

This makes the solution easier to follow and easier to check.

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Explaining the Method in Words

A clear explanation might be:

Since y is already isolated in the first equation, replace y in the second equation with its equivalent expression. Solve the resulting equation for x. Substitute that x-value back into either original equation to find y. Finally, check the ordered pair in both original equations.

Being able to explain the reasoning is as important as performing the calculation.

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Choosing Which Equation to Use for Substitution Back

Once one variable is known, either original equation can usually be used to find the other.

Suppose:

y = 4x − 7

3x + 2y = 19

and you find:

x = 3

The easiest equation is clearly:

y = 4x − 7

So:

y = 4(3) − 7

y = 5

Choose the equation that makes the arithmetic easiest.

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Avoid Unnecessary Work

Suppose:

x = 5 − y

7x + 3y = 23

There is no reason to rearrange the first equation again.

Simply substitute:

7(5 − y) + 3y = 23

Efficient algebra uses the information already available.

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Real-World Application: Ticket Sales

Adult tickets cost $10 and student tickets cost $6.

A total of 40 tickets were sold for $320.

Let:

a = number of adult tickets

s = number of student tickets

Number of tickets:

a + s = 40

Money collected:

10a + 6s = 320

From the first equation:

a = 40 − s

Substitute:

10(40 − s) + 6s = 320

Expand:

400 − 10s + 6s = 320

400 − 4s = 320

Subtract 400:

−4s = −80

Therefore:

s = 20

Then:

a = 40 − 20

a = 20

So:

20 adult tickets and 20 student tickets were sold.

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Real-World Application: Two Numbers

The sum of two numbers is 27.

One number is 5 greater than the other.

Let:

x = first number

y = second number

Equations:

x + y = 27

x = y + 5

Substitute:

(y + 5) + y = 27

Combine:

2y + 5 = 27

2y = 22

y = 11

Then:

x = 11 + 5

x = 16

The numbers are:

16 and 11

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Real-World Application: Perimeter

A rectangle has perimeter 46 cm.

Its length is 5 cm greater than its width.

Let:

l = length

w = width

Equations:

2l + 2w = 46

l = w + 5

Substitute:

2(w + 5) + 2w = 46

Expand:

2w + 10 + 2w = 46

4w + 10 = 46

4w = 36

w = 9

Then:

l = 9 + 5

l = 14

Therefore the rectangle is:

14 cm by 9 cm

Check:

2(14) + 2(9) = 46 cm

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Real-World Application: Science

Suppose two quantities satisfy:

F = 3m + 2

and:

F = 5m − 6

At what value of m do the two models predict the same force?

Since both expressions equal F:

3m + 2 = 5m − 6

Add 6:

3m + 8 = 5m

Subtract 3m:

8 = 2m

Therefore:

m = 4

Substitute:

F = 3(4) + 2

F = 14

So the models agree at:

m = 4, F = 14

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Real-World Application: Comparing Plans

Plan A costs:

C = 15 + 4x

Plan B costs:

C = 27 + 2x

where x represents the number of units used.

To find when the plans cost the same:

15 + 4x = 27 + 2x

Subtract 2x:

15 + 2x = 27

Subtract 15:

2x = 12

Therefore:

x = 6

Substitute:

C = 15 + 4(6)

C = 39

The plans cost the same when:

x = 6

and the cost is:

$39

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From Words to a System

When creating a system from a word problem:

1. Identify the unknown quantities.
2. Define variables.
3. Find two relationships.
4. Write one equation for each relationship.
5. Look for an equation that can easily be solved for one variable.
6. Use substitution.
7. Interpret the ordered pair in context.

The final answer should usually include words and units, not only an ordered pair.

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Substitution Versus Elimination

Substitution is usually a strong choice when:

y = ...

or:

x = ...

already appears.

For example:

y = 4x − 3

2x + 5y = 11

Substitution is natural.

But for:

3x + 2y = 12

5x − 2y = 8

elimination may be faster because +2y and −2y cancel immediately when the equations are added.

Both methods should produce the same solution.

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Substitution Versus Graphing

Graphing finds the intersection visually.

Substitution finds the intersection algebraically.

Substitution has several advantages:

• it can give exact answers
• it works well with fractional solutions
• it does not depend on graph-reading accuracy
• it provides an algebraic explanation

Graphing is useful for understanding what the solution means geometrically.

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Common Mistakes

Mistake: Substituting into the same equation used to isolate the variable.

After finding:

y = 2x + 3

use this expression in the other equation to create an equation with one variable.

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Mistake: Replacing only part of the variable.

If:

y = 2x − 5

then every occurrence of y must be replaced by the entire expression:

(2x − 5)

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Mistake: Forgetting parentheses.

For:

3x − y

and:

y = 2x − 5

write:

3x − (2x − 5)

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Mistake: Losing negative signs.

Remember:

−(2x − 5) = −2x + 5

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Mistake: Stopping after finding one variable.

A system solution normally requires values for both variables.

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Mistake: Writing the ordered pair backward.

If:

x = 3

and:

y = 7

the solution is:

(3, 7)

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Mistake: Checking only one equation.

A solution to a system must satisfy both equations.

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Mistake: Assuming all solutions are integers.

Systems may have fractions or decimals as solutions.

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Mistake: Assuming every system has exactly one solution.

A system may have:

• one solution
• no solution
• infinitely many solutions

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Mistake: Choosing an unnecessarily difficult variable to isolate.

Look first for a variable with coefficient 1 or −1.

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Key Terms

System of equations — two or more equations involving the same variables that are considered simultaneously.

Solution to a system — values that make every equation in the system true.

Substitution — replacing a variable with an equivalent value or algebraic expression.

Substitution method — method for solving a system by expressing one variable in terms of another and substituting that expression into another equation.

Isolate — rearrange an equation so that one variable appears alone on one side.

Equivalent expression — expression with the same value as another expression.

Ordered pair — pair written (x, y) that gives the values of two variables.

Intersection — point where two graphs meet.

Linear system — system consisting of linear equations.

Consistent system — system with at least one solution.

Inconsistent system — system with no solution.

Independent system — system whose equations represent different lines that intersect once.

Dependent system — system whose equations represent the same line and therefore have infinitely many solutions.

Verification — checking a proposed solution in the original equations.

No solution — result when no ordered pair satisfies every equation.

Infinitely many solutions — result when the equations describe the same relationship.

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Key Takeaways

• A system solution must satisfy all equations simultaneously.
• Substitution replaces one variable with an equivalent expression.
• Substitution is especially useful when a variable is already isolated.
• Variables with coefficients of 1 or −1 are often easy to isolate.
• The main process is Isolate → Substitute → Solve → Substitute Back → Verify.
• Substitute into the other equation so that only one variable remains.
• Put substituted expressions in parentheses, especially when subtraction or multiplication is involved.
• Solve the resulting one-variable equation carefully.
• After finding one variable, substitute back to find the other.
• Write a two-variable solution as an ordered pair (x, y).
• Always check the solution in both original equations.
• Substitution can produce integer, fractional, decimal, or negative solutions.
• If substitution produces a false statement, the system has no solution.
• If substitution produces an identity that is always true, the system has infinitely many solutions.
• One solution corresponds to two lines intersecting once.
• No solution corresponds to distinct parallel lines.
• Infinitely many solutions correspond to the same line.
• Both equations written as y = ... can be solved efficiently by setting their expressions equal.
• Substitution is useful for exact solutions that may be difficult to read from a graph.
• In word problems, define the variables before writing equations.
• The final solution to a practical problem should be interpreted in context.
• Substitution is one of several methods for solving systems; elimination may sometimes be more efficient.
• Choosing an efficient method is part of good algebraic problem solving.

Check Your Understanding

1. What is a system of equations?
2. What does it mean for an ordered pair to be a solution to a system?
3. Explain what substitution means in algebra.
4. Why is y = 3x + 2 especially convenient for the substitution method?
5. Solve: y = x + 4 and 2x + y = 13.
6. Solve: x = y + 2 and 3x + y = 14.
7. Solve: y = 2x − 3 and x + y = 9.
8. Solve: x + y = 12 and 2x + 3y = 29.
9. Solve: y = −x + 5 and 3x + y = 9.
10. Solve: y = x/2 + 2 and x + y = 8.
11. Explain why parentheses should be used when substituting an expression such as 2x − 5.
12. Solve: y = 3x + 1 and y = −2x + 11.
13. Verify whether (3, 4) solves x + y = 7 and 2x − y = 2.
14. Explain why a proposed solution must be checked in both equations.
15. What does a false statement such as 7 = 10 indicate after substitution?
16. What does a true statement such as 4 = 4 indicate after substitution?
17. Solve and classify the system: y = 2x + 1 and 2y = 4x + 2.
18. Solve and classify the system: y = 3x − 2 and 2y = 6x + 5.
19. Two numbers have a sum of 34. One number is 8 greater than the other. Create a system and solve it using substitution.
20. A rectangle has a perimeter of 54 cm. Its length is 7 cm greater than its width. Define variables, create a system, solve it using substitution, verify the solution, and interpret your answer.