Multiplying Before Elimination

Sometimes a system of equations is not immediately ready for elimination because none of the variable coefficients are equal or opposite.

For example:

2x + 3y = 12

x + 2y = 7

If we add or subtract these equations immediately, neither x nor y disappears.

Instead, we can multiply one or both equations by a carefully chosen number to create coefficients that will cancel.

For example, multiply the second equation by −2:

−2(x + 2y) = −2(7)

giving:

−2x − 4y = −14

Now:

2x + 3y = 12

−2x − 4y = −14

——————————

−y = −2

Therefore:

y = 2

Multiplication prepares the equations for elimination.

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Why Do We Multiply Before Eliminating?

The goal of elimination is to make one variable disappear.

This happens when its coefficients are:

equal and opposite

such as:

+5x and −5x

because:

5x + (−5x) = 0

or when equal coefficients are subtracted:

7y − 7y = 0

If the coefficients do not have this relationship, we can multiply equations to create it.

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Recognizing When Multiplication Is Needed

Consider:

3x + 2y = 14

5x − 4y = 2

The y-coefficients are:

2 and −4

They are not currently opposites.

However, multiplying the first equation by 2 gives:

6x + 4y = 28

Now the y terms:

+4y

and:

−4y

will cancel when the equations are added.

This is a strong sign that multiplication is useful.

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When Multiplication Is Not Needed

Consider:

3x + 5y = 17

2x − 5y = 8

The y-coefficients are already:

+5 and −5

Therefore, simply add the equations.

Multiplying first would create unnecessary work.

Always check whether a variable can already be eliminated before multiplying anything.

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Equal Coefficients

Consider:

4x + 3y = 16

4x − 2y = 6

The x-coefficients are equal.

Subtracting the equations eliminates x.

Again, no multiplication is necessary.

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The Main Strategy

For systems requiring multiplication, use:

Align → Inspect → Choose → Multiply → Eliminate → Solve → Substitute Back → Verify

Align

Write both equations in the same form.

Inspect

Compare the x- and y-coefficients.

Choose

Decide which variable will be easiest to eliminate.

Multiply

Multiply one or both equations to create equal or opposite coefficients.

Eliminate

Add or subtract the equations.

Solve

Solve for the remaining variable.

Substitute Back

Find the second variable.

Verify

Check the ordered pair in both original equations.

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Multiply the Entire Equation

This is one of the most important rules.

Suppose:

2x + 3y = 11

Multiply by 4.

Every term must be multiplied:

4(2x + 3y) = 4(11)

Therefore:

8x + 12y = 44

You cannot multiply only one term.

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Why the Entire Equation Must Be Multiplied

An equation represents equality.

If:

2x + 3y = 11

then multiplying both sides by 4 preserves equality:

4(2x + 3y) = 4(11)

But changing only one term would create a different equation that is no longer equivalent to the original.

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Equivalent Equations

These equations are equivalent:

x + 2y = 5

2x + 4y = 10

3x + 6y = 15

−4x − 8y = −20

Each is obtained by multiplying the entire original equation by a nonzero constant.

They all represent the same line.

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Multiplying One Equation

Solve:

3x + 2y = 14

5x − 4y = 2

The y-coefficients are:

2 and −4

Multiply the first equation by 2:

2(3x + 2y) = 2(14)

giving:

6x + 4y = 28

Now use the second equation:

5x − 4y = 2

Add:

6x + 4y = 28

5x − 4y = 2

——————————

11x = 30

Therefore:

x = 30/11

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Substitute Back

Use:

3x + 2y = 14

Substitute:

x = 30/11

3(30/11) + 2y = 14

90/11 + 2y = 154/11

2y = 64/11

Therefore:

y = 32/11

Solution:

(30/11, 32/11)

Fractional solutions are completely valid.

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Choosing the Easier Variable

Consider:

2x + 3y = 13

4x + 5y = 23

To eliminate x, multiply the first equation by −2:

−4x − 6y = −26

Then add:

4x + 5y = 23

——————————

−y = −3

So:

y = 3

Only one equation had to be multiplied.

This makes eliminating x an efficient choice.

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Finish the Example

Substitute y = 3:

2x + 3(3) = 13

2x + 9 = 13

2x = 4

Therefore:

x = 2

Solution:

(2, 3)

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Look for Coefficients That Are Multiples

Consider:

3x + 4y = 18

6x − 5y = 7

The x-coefficients are:

3 and 6

Since 6 is already a multiple of 3, multiply the first equation by −2:

−6x − 8y = −36

Now add:

6x − 5y = 7

——————————

−13y = −29

Only one equation needed to be changed.

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Use the Least Common Multiple

Consider:

4x + 3y = 10

6x − 5y = 8

The x-coefficients are:

4 and 6

Their least common multiple is:

12

We can create:

+12x

and:

−12x

Multiply the first equation by 3:

12x + 9y = 30

Multiply the second equation by −2:

−12x + 10y = −16

Now add.

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Continue the Example

Add:

12x + 9y = 30

−12x + 10y = −16

——————————

19y = 14

Therefore:

y = 14/19

Then substitute back to find x.

The important point is that the multipliers:

3 and −2

were chosen because they produced opposite x-coefficients.

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Why the LCM Is Useful

Suppose the coefficients are:

6 and 8

Possible common multiples include:

24, 48, 72, ...

The least common multiple is:

24

So we can multiply:

6 × 4 = 24

and:

8 × 3 = 24

Using the smallest convenient common multiple usually keeps the numbers easier to work with.

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You Do Not Always Need the LCM

The LCM is often useful, but any pair of multipliers that creates equal or opposite coefficients can work.

For example, coefficients:

2 and 3

could be changed to:

6 and −6

or:

12 and −12

or:

18 and −18

But:

6 and −6

usually produces the simplest arithmetic.

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Multiplying Both Equations

Solve:

2x + 3y = 12

5x + 2y = 19

Neither variable has coefficients that are simple multiples.

To eliminate y:

LCM of 3 and 2:

6

Multiply the first equation by 2:

4x + 6y = 24

Multiply the second equation by −3:

−15x − 6y = −57

Now add:

−11x = −33

Therefore:

x = 3

---

Complete the Solution

Substitute x = 3 into:

2x + 3y = 12

2(3) + 3y = 12

6 + 3y = 12

3y = 6

Therefore:

y = 2

Solution:

(3, 2)

---

Verify

First equation:

2(3) + 3(2) = 12

6 + 6 = 12 ✓

Second equation:

5(3) + 2(2) = 19

15 + 4 = 19 ✓

Therefore:

(3, 2)

is correct.

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Choosing Between x and y

Consider:

3x + 4y = 20

5x + 2y = 18

To eliminate y:

Multiply the second equation by −2:

−10x − 4y = −36

Only one equation needs multiplication.

To eliminate x:

We would need to create a common multiple of 3 and 5, requiring both equations to be multiplied.

Therefore eliminating y is probably more efficient.

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Efficiency Matters

Both x and y can usually be eliminated.

The best choice is normally the variable that requires:

• fewer multiplications
• smaller multipliers
• simpler arithmetic
• fewer fractions

Efficient choices reduce the chance of errors.

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Positive or Negative Multiplier?

Suppose:

2x + 3y = 11

5x + 6y = 25

To eliminate y, you could multiply the first equation by:

2

giving:

4x + 6y = 22

Then subtract the equations.

Or multiply it by:

−2

giving:

−4x − 6y = −22

Then add the equations.

Both methods work.

Choose whichever operation feels clearer.

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Addition Is Often Safer

Many students find it easier to create opposite coefficients and then add.

For example:

3x + 2y = 8

5x + 4y = 14

Multiply the first equation by −2:

−6x − 4y = −16

Now add:

−6x − 4y = −16

5x + 4y = 14

——————————

−x = −2

Therefore:

x = 2

This avoids subtracting an entire equation and reduces sign errors.

---

Subtraction Also Works

Using the same original system:

3x + 2y = 8

5x + 4y = 14

multiply the first equation by 2:

6x + 4y = 16

Now subtract the second equation:

(6x + 4y) − (5x + 4y) = 16 − 14

Therefore:

x = 2

Both approaches are valid.

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Be Careful with Negative Multipliers

Consider:

2x − 3y = 7

Multiply by −4:

−8x + 12y = −28

Notice:

−4(−3y) = +12y

Negative × negative = positive.

Sign errors are one of the most common mistakes when multiplying equations.

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Example with Negative Coefficients

Solve:

2x − 3y = −1

5x + 2y = 16

To eliminate y:

Multiply the first equation by 2:

4x − 6y = −2

Multiply the second equation by 3:

15x + 6y = 48

Add:

19x = 46

Therefore:

x = 46/19

The procedure is unchanged even when coefficients or constants are negative.

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Equations May Need Rearranging First

Consider:

y = 2x + 5

3x + 4y = 7

Before using elimination, rewrite the first equation in standard form:

−2x + y = 5

Now the system is:

−2x + y = 5

3x + 4y = 7

We can now compare coefficients easily.

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Prepare for Elimination

To eliminate y, multiply the first equation by −4:

8x − 4y = −20

Now add:

3x + 4y = 7

——————————

11x = −13

Therefore:

x = −13/11

Then substitute back to find y.

Rearranging equations into standard form often makes the required multiplication much easier to identify.

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Standard Form

Elimination is usually easiest when equations are written:

Ax + By = C

For example:

3x + 2y = 14

5x − 4y = 2

This arrangement places:

• x terms together
• y terms together
• constants together

and makes coefficient comparison straightforward.

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Fractions in a System

Sometimes equations themselves contain fractional coefficients.

For example:

x/2 + y/3 = 4

x − y = 1

Before elimination, it may be useful to clear the fractions.

The LCD of 2 and 3 is 6.

Multiply the entire first equation by 6:

3x + 2y = 24

Now the system is:

3x + 2y = 24

x − y = 1

The equations are much easier to work with.

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Clear Fractions Before Eliminating

For:

x/3 + y/2 = 5

multiply by 6:

2x + 3y = 30

This is an equivalent equation.

Clearing fractions can make the elimination process much cleaner.

---

Decimal Coefficients

Consider:

0.5x + 0.2y = 3

It may be useful to multiply the entire equation by 10:

5x + 2y = 30

This removes the decimals without changing the solutions.

The resulting equation is often easier to combine with another equation.

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Multiplication Can Simplify Before Elimination

Multiplication is therefore useful for more than matching coefficients.

It can also:

• clear fractions
• clear decimals
• create integer coefficients
• make equations easier to compare

The goal is always to create an equivalent system that is easier to solve.

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No Solution After Multiplication

Consider:

2x + 3y = 7

4x + 6y = 20

Multiply the first equation by −2:

−4x − 6y = −14

Add:

4x + 6y = 20

——————————

0 = 6

This is false.

Therefore:

No solution

The equations represent parallel lines.

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Infinitely Many Solutions After Multiplication

Consider:

2x + 3y = 7

4x + 6y = 14

Multiply the first equation by −2:

−4x − 6y = −14

Add:

4x + 6y = 14

——————————

0 = 0

This statement is always true.

Therefore:

Infinitely many solutions

The equations represent the same line.

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Multiplication Can Reveal the Relationship

Notice the difference:

2x + 3y = 7

and:

4x + 6y = 14

The entire second equation is exactly twice the first.

Therefore the equations describe the same relationship.

But:

2x + 3y = 7

and:

4x + 6y = 20

have proportional variable coefficients but inconsistent constants.

Therefore the lines are parallel.

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Verification Is Essential

Suppose your solution is:

(3, 2)

for:

2x + 3y = 12

5x + 2y = 19

Check the first equation:

2(3) + 3(2) = 12

12 = 12 ✓

Check the second:

5(3) + 2(2) = 19

19 = 19 ✓

Therefore the solution is correct.

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Why Verify in the Original Equations?

You may have multiplied the equations several times while solving.

Checking the original equations is best because it confirms that:

• the multiplication was correct
• the elimination was correct
• the arithmetic was correct
• the final ordered pair solves the actual original system

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Catching a Multiplication Error

Suppose:

3x − 2y = 8

is multiplied by 4.

Correct:

12x − 8y = 32

If a student writes:

12x − 2y = 32

the equation is no longer equivalent.

Verification in the original system will often reveal this error.

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Real-World Example: Ticket Sales

Adult tickets cost $14.

Student tickets cost $9.

A total of 80 tickets are sold for $900.

Let:

a = adult tickets

s = student tickets

System:

a + s = 80

14a + 9s = 900

To eliminate s, multiply the first equation by −9:

−9a − 9s = −720

Add:

14a + 9s = 900

——————————

5a = 180

Therefore:

a = 36

Then:

s = 80 − 36

s = 44

Therefore:

36 adult tickets and 44 student tickets were sold.

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Real-World Example: Shopping

Two notebooks and three pens cost $12.

Three notebooks and two pens cost $13.

Let:

n = notebook price

p = pen price

System:

2n + 3p = 12

3n + 2p = 13

To eliminate n, the LCM of 2 and 3 is 6.

Multiply the first equation by 3:

6n + 9p = 36

Multiply the second equation by −2:

−6n − 4p = −26

Add:

5p = 10

Therefore:

p = 2

Substitute:

2n + 3(2) = 12

2n = 6

n = 3

Therefore:

• notebook = $3
• pen = $2

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Real-World Example: Two Types of Items

A store sells small packages and large packages.

Three small packages and four large packages weigh 26 kg.

Five small packages and two large packages weigh 28 kg.

Let:

s = mass of one small package

l = mass of one large package

System:

3s + 4l = 26

5s + 2l = 28

To eliminate l, multiply the second equation by −2:

−10s − 4l = −56

Add:

3s + 4l = 26

——————————

−7s = −30

Therefore:

s = 30/7 kg

The fractional result is mathematically acceptable even if it is less convenient.

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Comparing Multiplication Choices

Suppose:

6x + 5y = 20

4x − 3y = 8

To eliminate x:

LCM of 6 and 4:

12

Multipliers:

2 and −3

To eliminate y:

LCM of 5 and 3:

15

Multipliers:

3 and 5

Both work.

The x route involves somewhat smaller coefficients, so eliminating x may be preferable.

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Do Not Multiply More Than Necessary

Suppose:

3x + 4y = 10

6x − 5y = 8

The x-coefficients are already related by a factor of 2.

Multiply only the first equation by −2:

−6x − 8y = −20

There is no reason to multiply both equations.

Efficient solving means doing only the algebra that is necessary.

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A Challenging Example

Solve:

4x + 7y = 25

6x − 5y = 3

To eliminate x:

LCM of 4 and 6:

12

Multiply the first equation by 3:

12x + 21y = 75

Multiply the second by −2:

−12x + 10y = −6

Add:

31y = 69

Therefore:

y = 69/31

Substitute into either original equation to determine x.

Even when the arithmetic becomes complicated, the strategy does not change.

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Explaining Why Multiplication Works

Suppose:

2x + y = 5

Multiplying by 3 gives:

6x + 3y = 15

These equations have exactly the same solutions because the second equation simply states three times the same equality.

Multiplication changes the appearance of the equation but not the line or its solution set.

This allows us to deliberately create coefficients that cancel with coefficients in another equation.

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Elimination as Adding Equivalent Equations

Suppose:

2x + 3y = 12

and:

5x + 2y = 19

We multiply them to create:

4x + 6y = 24

−15x − 6y = −57

These are equivalent to the original equations.

Adding them gives:

−11x = −33

Every common solution of the original system must also satisfy this new equation.

That is the algebraic reason elimination works.

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A Useful Decision Process

Before multiplying, ask:

Are any coefficients already opposite?

If yes → add.

Are any coefficients already equal?

If yes → subtract.

Is one coefficient a simple multiple of another?

If yes → multiply one equation.

Otherwise, can I use an LCM?

If yes → multiply both equations.

This decision process can make elimination much faster.

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Common Mistakes

Mistake: Multiplying only the variable you want to eliminate.

If multiplying an equation, multiply every term.

---

Mistake: Forgetting the constant.

For:

2x + 3y = 8

multiplied by 4:

8x + 12y = 32

not:

8x + 12y = 8

---

Mistake: Losing signs with negative multipliers.

For:

3x − 2y = 5

multiplied by −3:

−9x + 6y = −15

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Mistake: Multiplying when elimination is already possible.

Always inspect the coefficients first.

---

Mistake: Choosing unnecessarily large multipliers.

Use small convenient multiples, usually based on the LCM.

---

Mistake: Multiplying both equations when only one needs changing.

Look for coefficients that are already multiples.

---

Mistake: Creating equal coefficients and then adding them.

If you create:

+6x and +6x

subtract.

If you want to add, create:

+6x and −6x.

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Mistake: Creating opposite coefficients and then subtracting them.

Opposite coefficients cancel by addition.

---

Mistake: Forgetting to substitute back.

Elimination initially gives only one variable.

---

Mistake: Failing to verify.

Check the final ordered pair in both original equations.

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Key Terms

Elimination — method for solving a system by combining equations so one variable disappears.

Coefficient — numerical factor multiplying a variable.

Multiplier — number by which an entire equation is multiplied.

Equivalent equation — equation having the same solution set as another equation.

Opposite coefficients — coefficients with equal magnitude and opposite signs, such as +6 and −6.

Equal coefficients — coefficients having the same value and sign.

Least common multiple (LCM) — smallest positive number that is a multiple of two or more numbers.

Standard form — linear equation written as Ax + By = C.

System of equations — two or more equations involving the same variables.

Solution to a system — values satisfying every equation simultaneously.

Ordered pair — pair (x, y) representing values of the variables.

Substitute back — use a known variable value in an original equation to determine another variable.

Verification — checking a solution in the original equations.

Equivalent system — system transformed without changing its solutions.

No solution — situation in which no ordered pair satisfies both equations.

Infinitely many solutions — situation in which both equations represent the same relationship.

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Key Takeaways

• Multiplication is used before elimination when coefficients are not already suitable for cancellation.
• Always check whether multiplication is actually necessary.
• Equal coefficients can be eliminated by subtraction.
• Opposite coefficients can be eliminated by addition.
• If one coefficient is a simple multiple of another, usually only one equation needs to be multiplied.
• Otherwise, use a common multiple to prepare both equations.
• The LCM often gives the smallest useful multipliers.
• Multiplying an equation means multiplying every term on both sides.
• Multiplying an equation by a nonzero constant creates an equivalent equation.
• Equivalent equations represent the same line and have the same solutions.
• Negative multipliers can be useful for creating opposite coefficients.
• Be especially careful with signs when using negative multipliers.
• Choose the variable that requires the simplest multiplication.
• Smaller multipliers generally reduce arithmetic errors.
• Fractions can often be cleared before elimination by multiplying by an LCD.
• Decimals can often be removed by multiplying by a power of 10.
• Once one variable is eliminated, solve for the remaining variable.
• Substitute back into an original equation to find the second variable.
• Write the solution as an ordered pair (x, y).
• Verify the answer in both original equations.
• A false statement such as 0 = 6 means there is no solution.
• A true statement such as 0 = 0 means there are infinitely many solutions.
• Multiplying before elimination is not a separate method; it is a preparation step that makes elimination possible.
• Good elimination involves choosing efficient multipliers rather than simply multiplying equations randomly.

Check Your Understanding

1. Explain why multiplication is sometimes necessary before elimination.
2. Which variable can be eliminated immediately in 3x + 4y = 12 and 5x − 4y = 8?
3. Does 2x + 3y = 7 and 4x − 5y = 11 require multiplication before elimination? Explain.
4. Multiply 3x + 2y = 8 by 4.
5. Multiply 2x − 5y = 7 by −3.
6. Explain why every term in an equation must be multiplied by the same multiplier.
7. Solve 2x + 3y = 13 and 4x + 5y = 23 using multiplication before elimination.
8. Solve 3x + 2y = 14 and 5x − 4y = 2.
9. Solve 2x + 3y = 12 and 5x + 2y = 19 by multiplying both equations.
10. For 4x + 3y = 10 and 6x − 5y = 8, determine useful multipliers for eliminating x.
11. For the same system, determine useful multipliers for eliminating y.
12. Explain which variable you would choose to eliminate in 3x + 4y = 20 and 5x + 2y = 18, and why.
13. Rewrite x/2 + y/3 = 4 without fractional coefficients.
14. Rewrite 0.4x + 0.3y = 2.5 without decimal coefficients.
15. Explain why using the LCM can make elimination more efficient.
16. Solve and classify 2x + 3y = 7 and 4x + 6y = 14.
17. Solve and classify 2x + 3y = 7 and 4x + 6y = 20.
18. Two notebooks and three pens cost $12. Three notebooks and two pens cost $13. Create a system and solve it using multiplication before elimination.
19. Adult tickets cost $14 and student tickets cost $9. A school sells 80 tickets for $900. Create and solve a system using elimination, then verify your answer.
20. Explain the complete Align → Inspect → Choose → Multiply → Eliminate → Solve → Substitute Back → Verify process and explain why multiplying an equation does not change its solutions.