Elimination after multiplying

Often neither unknown has coefficients of the same size. Then multiply one equation, or both, by suitable numbers so that one unknown does, and eliminate as before.

Why it is allowed: multiplying every term of an equation by the same non-zero number gives an equivalent equation with the same solutions. For example, \(x+5y=7\) and \(3x+15y=21\) are satisfied by exactly the same pairs, such as \((2,1)\) and \((7,0)\).

Method

1. Choose the unknown to eliminate. Prefer one whose coefficients are already factors of each other, or whose signs are opposite so that you can add.
2. Multiply one equation (or both) so that those coefficients have the same size. The target is usually the lowest common multiple of the two coefficients.
3. Multiply every term, including the constant on the right-hand side, and label the new equations.
4. Add or subtract to eliminate the unknown, solve, substitute back and check in both original equations.

Example

Multiplying one equation. Solve \(7x+2y=24\) \((1)\) and \(5x-4y=-10\) \((2)\).

\(\begin{aligned} (1)\times2:\quad 14x+4y&=48\quad(3) \\ (3)+(2):\quad 19x&=38 \\ x&=2 \end{aligned}\)

Substitute into \((1)\): \(14+2y=24\), so \(y=5\). Check in \((2)\): \(5(2)-4(5)=10-20=-10\).

Multiplying both equations. Solve \(3x+7y=-1\) \((1)\) and \(5x-3y=13\) \((2)\). To eliminate \(y\), use the LCM of \(7\) and \(3\), which is \(21\).

\(\begin{aligned} (1)\times3:\quad 9x+21y&=-3\quad(3) \\ (2)\times7:\quad 35x-21y&=91\quad(4) \\ (3)+(4):\quad 44x&=88 \\ x&=2 \end{aligned}\)

Substitute into \((1)\): \(6+7y=-1\), so \(y=-1\). Check in \((2)\): \(5(2)-3(-1)=10+3=13\).

Check: a common error is to multiply only the left-hand side, writing \(14x+4y=24\) instead of \(14x+4y=48\). Always check in the original equations, not the multiplied ones.