Grade by Grade
Analytic Geometry · Analytic Geometry

Solving Linear Systems by Elimination

1.1About 70 minutes
Learning goals
  • I can explain why adding or subtracting equations keeps the same solution.
  • I can solve a linear system by elimination, multiplying one or both equations when needed.
  • I can clear fractions and decimals before eliminating.
  • I can recognize no solution and infinitely many solutions during elimination.

Words to know

elimination
Adding or subtracting equations to remove one variable.
coefficient
The number that multiplies a variable, such as 3 in 3x.
equivalent system
A system with exactly the same solution as another system.
lowest common multiple (LCM)
The smallest positive number that two numbers both divide into evenly.
opposites
Numbers with the same size and opposite signs, such as 4 and −4.

A different way to make a variable disappear

Suppose a café sells a muffin and a hot chocolate for $6, and two muffins and a hot chocolate for $8.50. You can probably see the answer quickly: the second order has one extra muffin and costs $2.50 more, so a muffin costs $2.50 and the hot chocolate costs $3.50.

What you just did was subtract one equation from the other to make a variable disappear. That is the elimination method. It is often the fastest method when both equations are in standard form, ax + by = c.

Why it works

If a = b and c = d, then a + c = b + d. Adding (or subtracting) equal amounts to both sides keeps an equation balanced. So you may add or subtract two equations and the solution of the system does not change.

You may also multiply every term of an equation by the same non-zero number. The new equation describes the same line, so it is an equivalent equation. A system made of equivalent equations is an equivalent system: it has the same solution.

The method of elimination

  1. Write both equations in the form ax + by = c, with like terms lined up.
  2. If needed, multiply one or both equations so that one variable has equal or opposite coefficients.
  3. Add the equations if the coefficients are opposites (like +2y and −2y). Subtract if they are the same (like 2x and 2x).
  4. Solve for the remaining variable.
  5. Substitute back into either original equation to find the other variable.
  6. Check in both original equations.
3x + 2y = 165x − 2y = 0+8x + 0y = 16add the left sidesand the right sides+2y and −2y cancel:y is eliminated, x = 2
Adding the two equations of Example 1. The coefficients +2y and −2y are opposites, so adding eliminates y and leaves 8x = 16.
Worked example

Example 1: Add to eliminate

Solve by elimination: 3x + 2y = 16 and 5x − 2y = 0.

  1. The y-coefficients are +2 and −2: opposites. Add the equations: (3x + 5x) + (2y − 2y) = 16 + 0, so 8x = 16.
  2. Solve: x = 2.
  3. Substitute into the first equation: 3(2) + 2y = 16, so 6 + 2y = 16, 2y = 10, y = 5.
  4. Check: 3(2) + 2(5) = 6 + 10 = 16 ✓ and 5(2) − 2(5) = 10 − 10 = 0 ✓.

The solution is (2, 5).

−1123456123456780xy(2, 5)3x + 2y = 165x − 2y = 0x = 2
The two original lines cross at (2, 5). Eliminating y gives the vertical line x = 2, which passes through the same crossing point. Elimination swaps a hard pair of lines for an easy one without moving the solution.
Worked example

Example 2: Subtract to eliminate

Solve: 2x + 3y = 13 and 2x − y = 1.

  1. The x-coefficients are both 2: the same. Subtract the second equation from the first.
  2. (2x − 2x) + (3y − (−y)) = 13 − 1, so 0x + 4y = 12. Careful: 3y − (−y) = 3y + y = 4y.
  3. Solve: y = 3.
  4. Substitute into the second equation: 2x − 3 = 1, so 2x = 4, x = 2.
  5. Check: 2(2) + 3(3) = 4 + 9 = 13 ✓ and 2(2) − 3 = 1 ✓.

The solution is (2, 3).

Add or subtract?

Opposite coefficients (+5 and −5): add. Same coefficients (+5 and +5, or −5 and −5): subtract. When you subtract, subtract every term, including the constant on the right side. Many students prefer to multiply one equation by −1 and then always add; that is fine too.

Worked example

Example 3: Multiply both equations first

Solve: 3x + 4y = −2 and 2x − 3y = 10.

  1. No coefficients match. To eliminate x, make both x-coefficients 6 (the lowest common multiple of 3 and 2).
  2. Multiply the first equation by 2: 6x + 8y = −4. Multiply the second by 3: 6x − 9y = 30.
  3. Subtract the second new equation from the first: 8y − (−9y) = −4 − 30, so 17y = −34, y = −2.
  4. Substitute into 3x + 4y = −2: 3x − 8 = −2, 3x = 6, x = 2.
  5. Check: 3(2) + 4(−2) = 6 − 8 = −2 ✓ and 2(2) − 3(−2) = 4 + 6 = 10 ✓.

The solution is (2, −2). (You could also eliminate y by multiplying by 3 and 4 and then adding: 9x + 12y = −6 and 8x − 12y = 40 give 17x = 34.)

Worked example

Example 4: Decimals

Solve: 0.5x + 0.2y = 3.1 and 0.3x − 0.4y = 0.3.

  1. Multiply both equations by 10 to clear the decimals: 5x + 2y = 31 and 3x − 4y = 3.
  2. Multiply the first new equation by 2: 10x + 4y = 62.
  3. Add it to 3x − 4y = 3: 13x = 65, so x = 5.
  4. Substitute into 5x + 2y = 31: 25 + 2y = 31, 2y = 6, y = 3.
  5. Check in the original equations: 0.5(5) + 0.2(3) = 2.5 + 0.6 = 3.1 ✓ and 0.3(5) − 0.4(3) = 1.5 − 1.2 = 0.3 ✓.

The solution is (5, 3).

Special cases

As with substitution, if both variables are eliminated:

  • a false statement (such as 0 = 9) means no solution;
  • a true statement (such as 0 = 0) means infinitely many solutions.

For example, 2x + 3y = 5 and 4x + 6y = 19: multiply the first by 2 to get 4x + 6y = 10, then subtract: 0 = −9. False, so there is no solution.

Just for fun

Funny-but-true: elimination is ancient. A Chinese book called The Nine Chapters on the Mathematical Art, about 2000 years old, solved systems by setting up counting rods in columns and subtracting columns from each other. Your calculator is just a very fast stack of counting rods.

Hands-on activity

The secret snack prices

You need: two snack items from the kitchen (for example, a granola bar and a juice box), paper, a pencil and a grown-up.

  1. The grown-up secretly chooses a price for each item (whole dollars or 25-cent steps keep it simple) and writes them down without showing you.
  2. The grown-up makes two "orders" and tells you only the total of each. For example: "3 bars and 2 juices cost $7.50. 2 bars and 5 juices cost $10.50."
  3. Let b be the price of a bar and j the price of a juice. Write the system: 3b + 2j = 7.50 and 2b + 5j = 10.50.
  4. Solve by elimination. (Here: multiply by 2 and 3 to get 6b + 4j = 15 and 6b + 15j = 31.50; subtract to get 11j = 16.50, so j = $1.50, then b = $1.50.)
  5. Compare with the secret prices. Then make up a system for the grown-up to solve.

What to look for: a clear choice of which variable to eliminate, correct multiplying of every term, and a check in both orders.

Your turn

1

Solve by elimination: x + y = 9 and x − y = 3.

2

Solve: 4x + 3y = 11 and 2x + 3y = 7.

3

Solve: 3x − 2y = 4 and x + 2y = 12.

4

Solve: 2x + 5y = 1 and 3x − y = 10.

5

Solve: 5x + 2y = 4 and 3x + 5y = −9.

6

Solve: 0.2x + 0.3y = 1.3 and 0.5x − 0.1y = 0.7.

7

Show that 3x − y = 2 and 6x − 2y = 9 has no solution.

More practice: warm-up

1

Which variable is easiest to eliminate in 2x + 7y = 3 and 5x − 7y = 11? Add or subtract?

2

Which variable is easiest to eliminate in 4x + y = 6 and 4x − 3y = 2? Add or subtract?

3

Multiply every term of 2x − 3y = 5 by 4.

4

Simplify 5y − (−2y).

5

Solve: x + y = 7 and −x + y = 1.

More practice: on your own

1

Solve: 3x + 2y = 7 and 5x − 4y = 41.

2

Solve: 4x − 3y = 18 and 6x + 5y = −11.

3

Solve: x/3 + y/2 = 5 and x/2 − y/4 = 1.

4

Solve: 7x + 2y = 10 and 3x = 4y + 14. (Rearrange first.)

5

At a hockey arena, 4 hot dogs and 3 drinks cost $25.50, and 2 hot dogs and 5 drinks cost $21.50. Find the price of each.

Challenge

1

Solve: 2(x + y) − 3(x − y) = 9 and 4x − (y − 2) = 4.

2

Find a and b if the system ax + by = 7 and bx + ay = 8 has solution (2, 1).

3

Explain why multiplying an equation by 0 is not allowed in the elimination method.

Check your understanding

Quiz

1.

To solve 4x + 3y = 10 and 2x − 3y = 8 by elimination, the best first step is to:

2.

Solve 5x + 2y = 9 and 5x − y = 3.

3.

Which equation is equivalent to 3x − 2y = 5?

4.

You want to eliminate y from 2x + 3y = 4 and 5x + 2y = −1. What should you multiply by?

5.

Elimination gives 0 = 0. The system has:

6.

In 2x + 3y = 13 and 2x − y = 1, after subtracting the second equation from the first you get:

For parents and teachers

Big idea: adding or subtracting equations (after multiplying if needed) removes one variable but keeps the same solution. This lesson teaches expectation 1.1 (solving systems by elimination).

How to help: Ask your teen why they chose to add or subtract, and which variable they eliminated. Encourage neat, lined-up work: x under x, y under y, constants under constants. Practise with real price puzzles (the snack activity).

Common difficulties:

  • Multiplying only the left side, or only one term, instead of every term.
  • Sign errors when subtracting: 3y − (−y) is 4y, and 13 − (−5) is 18.
  • Adding when they should subtract (or the reverse). The rule: opposites add, sames subtract.
  • Real data do not always give whole-number answers; fractions and decimals are fine if they check.