Solve a pair of simultaneous linear equations by elimination

First equation

\(\displaystyle 2x+y=13\)

Second equation

\(\displaystyle 2x+3y=23\)

Subtract the first equation from the second. Which equation remains?

Subtract every term of the first equation from the matching term of the second

First equation

\(\displaystyle 2x+4y=20\)

Second equation

\(\displaystyle 3x-2y=6\)

Multiplier for the second equation

\(\displaystyle k\)

The table gives the coefficients of the unknowns and the right-hand side. The first two rows are the original equations. The next row multiplies the second equation by the selected multiplier. The final row adds the first and scaled equations. Compare the coefficient of the second unknown at the two requested settings

What happens to the coefficient of y in the sum row when the multiplier moves from two to three?

Make one pair of terms cancel

Use equal coefficients to subtract, or opposite coefficients to add. Their sizes must match. If needed, multiply every term on both sides of one equation first

1. 1Scale a whole equation Double the second equation so the coefficients of y have equal sizes and opposite signs \(\displaystyle \begin{aligned}2(x-y)&=2(1)\\2x-2y&=2\end{aligned}\)
2. 2Add both sides The terms in y cancel. Keep the other terms and both constants \(\displaystyle \begin{gathered}(3x+2y)+(2x-2y)\\=23+2\\5x=25\\x=5\end{gathered}\)
3. 3Substitute back Use the simpler original equation to find the other unknown \(\displaystyle \begin{aligned}5-y&=1\\-y&=-4\\y&=4\end{aligned}\)
4. 4Check both originals Both values must make both original equations true \(\displaystyle \begin{aligned}3(5)+2(4)&=23\\5-4&=1\end{aligned}\)
5. 5When both equations need scaling If one multiplier cannot make a match, multiply both equations. Use the smallest common multiple of the chosen coefficient sizes. Then add or subtract as before. The worked example shows this method

First equation

\(\displaystyle 2x+3y=19\)

Second equation

\(\displaystyle 4x-y=17\)

Use elimination to find x

Make the coefficients of y equal in size and opposite in sign. Multiply both sides of the equation you change

First equation

\(\displaystyle 2x+3y=19\)

Second equation

\(\displaystyle 4x-y=17\)

Known value

\(\displaystyle x=5\)

Find y by substitution

Use the second equation, where y has a coefficient of negative one

What went wrong in the student's last line?

Compare each term in the bracket with its term in the last line