Simultaneous equations

Simultaneous equations are two equations that are both true at once. Solving them means finding the values of x and y that fit both.

To eliminate one letter:

1. Line up the equations so the x's, y's and numbers are in columns.
2. If one letter has the same coefficient in both, add or subtract to get rid of it.
3. Solve for the other letter.
4. Substitute back into either equation to find the first.

Example: solve 3x + y = 11 and 2x − y = 4.

1. The y's have opposite signs, so add: 5x = 15
2. x = 3
3. Substitute into 3x + y = 11: 9 + y = 11, so y = 2
4. x = 3, y = 2

Check in the other equation: 2 × 3 − 2 = 4.

If no letter has matching coefficients, multiply one or both equations so that one does.

Example: solve 2x + 3y = 12 and 5x − 2y = 11.

1. Multiply the first by 2: 4x + 6y = 24
2. Multiply the second by 3: 15x − 6y = 33
3. The y's now match with opposite signs, so add: 19x = 57
4. x = 3
5. Substitute into 2x + 3y = 12: 6 + 3y = 12, so y = 2
6. x = 3, y = 2

Check: 5 × 3 − 2 × 2 = 15 − 4 = 11.

Multiply every term, including the number on the right-hand side.

If one equation already says y = … (or x = …), substitute it into the other.

1. Replace that letter in the other equation with its expression, in brackets.
2. Solve the equation you get.
3. Substitute back to find the other letter.

Example: solve y = 2x − 1 and 3x + 2y = 19.

1. 3x + 2(2x − 1) = 19
2. 3x + 4x − 2 = 19, so 7x = 21 and x = 3
3. y = 2 × 3 − 1 = 5
4. x = 3, y = 5

Check: 3 × 3 + 2 × 5 = 9 + 10 = 19.

Word problems with two unknowns need two equations.

1. Choose a letter for each unknown.
2. Write one equation for each fact.
3. Solve, and answer the question in words, with units.

Example: 3 adult tickets and 2 child tickets cost £34. 1 adult ticket and 4 child tickets cost £28. Find the price of each ticket.

1. Let an adult ticket cost £a and a child ticket £c.
2. 3a + 2c = 34 and a + 4c = 28
3. Multiply the second by 3: 3a + 12c = 84
4. Subtract the first: 10c = 50, so c = 5
5. a + 4 × 5 = 28, so a = 8
6. Adult £8, child £5

Check: 3 × 8 + 2 × 5 = 24 + 10 = 34.

If one equation has an x² (or is a circle, x² + y² = r²), substitute the linear one into it. There are usually two pairs of answers.

1. Write the linear equation as y = … and substitute it.
2. Rearrange to = 0 and solve the quadratic.
3. Find the matching y for each x.

Example: solve x² + y² = 25 and y = x + 1.

1. x² + (x + 1)² = 25, so 2x² + 2x + 1 = 25
2. 2x² + 2x − 24 = 0, so x² + x − 12 = 0
3. (x + 4)(x − 3) = 0, so x = 3 or x = −4
4. x = 3, y = 4 or x = −4, y = −3