Simultaneous Equations — Elimination Method

Junior Cycle revision — learn the general strategy, follow the example, then practise.

1. General Strategy

Suppose we have two equations in two unknowns \(x\) and \(y\):

\[ \begin{cases} ax + by = c, \\ dx + ey = f. \end{cases} \]

Align coefficients. Decide whether to eliminate \(x\) or \(y\). Multiply one or both equations by suitable numbers so that the chosen variable has the same size coefficients with different signs in both equations.
Add Add the equations to eliminate the chosen variable. You now have one equation in a single variable.
Solve for the remaining variable. Solve this new single-variable equation.
Back-substitute. Substitute the found value into one of the original equations to find the other variable.

Tip: Always check your answers by substituting both values into the original equations.

Solve: \(\;2x + 3y = 1\) and \(\;x - y = 3\)

Step 1: Eliminate \(x\)

Multiply the second equation by −2:

\(2x + 3y = 1\)

\(-2x + 2y = -6\)

\(5y = -5 \Rightarrow y = -1\)

Step 2: Substitute \(y = -1\)

In \(x - y = 3\): \(x - (-1) = 3 \Rightarrow x = 2.\)

Check: \(2(2)+3(-1)=1\;\checkmark\) and \(2-(-1)=3\;\checkmark\)

Solution: \(\boxed{x=2,\;y=-1}\)

1. \(\;2x + y = 9,\quad -2x + 3y = 3\)

Add the equations:

\(2x + y = 9\)

\(-2x + 3y = 3\)

\(4y = 12 \Rightarrow y = 3\)

Substitute \(y = 3\) into \(2x + y = 9\): \(2x + 3 = 9 \Rightarrow x = 3.\)

Answer: \(\boxed{x=3,\;y=3}\)

2. \(\;3x - y = 5,\quad 3x + 2y = 17\)

Subtract the first from the second:

\(3x + 2y = 17\)

\(-(3x - y = 5)\)

\(3y = 12 \Rightarrow y = 4\)

Then \(3x - y = 5 \Rightarrow 3x - 4 = 5 \Rightarrow x = 3.\)

Answer: \(\boxed{x=3,\;y=4}\)

3. \(\;5x - 3y = 11,\quad 3x + 3y = 9\)

Add the equations:

\(5x - 3y = 11\)

\(+\,3x + 3y = 9\)

\(8x = 20 \Rightarrow x = 2.5\)

Substitute \(x = 2.5\) into \(3x + 3y = 9\): \(7.5 + 3y = 9 \Rightarrow y = 0.5.\)

Answer: \(\boxed{x=2.5,\;y=0.5}\)