Two unknowns need two equations. Solve them by elimination or substitution, set them up from a problem, and (Extended) solve a linear equation together with a quadratic one.
Make the coefficients of one unknown the same size, then add the equations if the signs differ or subtract if they are the same. Solve for the other unknown, substitute back, and check both values in both equations. The solution is where the two graphs cross.
Fig. 1: the graphs of 2x + y = 7 and x - y = 2 cross at (3, 1), the one pair of values that makes both equations true
Section 01
Eliminate one letter.
In the lab choose Two lines and set the sliders to the equations \(2x + y = 7\) and \(x - y = 2\). Watch the working add the equations to remove \(y\), and see the crossing point on the graph match the answer. Then change one coefficient and watch both the algebra and the crossing point move.
Worked example Add to eliminate
Solve \(2x + y = 7\) and \(x - y = 2\).
The \(y\) terms are \(+y\) and \(-y\): different signs, so add the equations.
\(3x = 9\), so \(x = 3\).
Substitute into \(x - y = 2\): \(3 - y = 2\), so \(y = 1\).
Check in the first: \(2(3) + 1 = 7\). Correct.
Answer: \(x = 3,\ y = 1\)
Worked example Multiply first
Solve \(3x + 2y = 12\) and \(5x - 3y = 1\).
Make the \(y\) coefficients both 6: multiply the first by 3 and the second by 2.
\(9x + 6y = 36\) and \(10x - 6y = 2\).
Signs differ, so add: \(19x = 38\), \(x = 2\).
Substitute: \(3(2) + 2y = 12\), so \(2y = 6\) and \(y = 3\).
Answer: \(x = 2,\ y = 3\)
Worked example Substitution
Solve \(y = 2x - 1\) and \(3x + 2y = 19\).
Replace \(y\) in the second equation: \(3x + 2(2x - 1) = 19\).
\(3x + 4x - 2 = 19\), so \(7x = 21\) and \(x = 3\).
\(y = 2(3) - 1 = 5\).
Answer: \(x = 3,\ y = 5\)
Your turn
Solve \(4x + 3y = 25\) and \(2x + 5y = 23\).
Double the second: \(4x + 10y = 46\). Same signs, so subtract the first: \(7y = 21\), \(y = 3\). Then \(2x + 15 = 23\), \(x = 4\).
Examiner trap
Same Signs Subtract. When you subtract \(2x - y = 5\) from \(2x + 3y = 17\), the \(y\) terms give \(3y - (-y) = 4y\), not \(2y\).
Section 02
Setting them up.
Choose a letter for each unknown, then turn each sentence of information into one equation. Two unknowns need two separate facts.
Worked example Ticket prices
2 adult tickets and 3 child tickets cost $31. 1 adult ticket and 4 child tickets cost $28. Find the cost of each ticket.
Let an adult ticket cost \(a\) and a child ticket cost \(c\) dollars.
\(2a + 3c = 31\) and \(a + 4c = 28\).
Double the second: \(2a + 8c = 56\). Subtract the first: \(5c = 25\), so \(c = 5\).
\(a + 20 = 28\), so \(a = 8\).
Answer: Adult $8, child $5
Your turn
5 pens and 2 rulers cost $3.70. 3 pens and 4 rulers cost $3.90. Find the cost of one pen.
\(5p + 2r = 3.70\) and \(3p + 4r = 3.90\). Double the first: \(10p + 4r = 7.40\). Subtract the second: \(7p = 3.50\), so \(p = 0.50\). A pen costs $0.50 (and a ruler $0.60).
Examiner trap
Write down what each letter stands for, with units. Many students find the right numbers and then give the child price when the question asked for the adult price.
Stage 1 · Learn
Check your understanding
Four quick checks. Each correct answer earns XP and records this skill on your profile.
Quick check+10 XP
Solve \(x + y = 10\) and \(x - y = 4\).
Quick check+10 XP
To eliminate \(y\) from \(3x + 2y = 11\) and \(5x - 2y = 13\), you should:
Quick check+10 XP
Solve \(y = 2x\) and \(x + y = 9\).
Quick check+10 XP
On a graph, the solution of two simultaneous linear equations is:
Section 03
A line and a curve (Extended).
When one equation is quadratic, substitution is the only way. Rearrange the linear equation to give \(y\) (or \(x\)), substitute it into the other, and solve the quadratic. A line can meet a curve twice, so expect two pairs of answers.
Worked example Line and parabola (Extended)
Solve \(y = x + 3\) and \(y = x^{2} + 1\).
Both equal \(y\): \(x^{2} + 1 = x + 3\).
\(x^{2} - x - 2 = 0\), so \((x - 2)(x + 1) = 0\).
\(x = 2\) or \(x = -1\).
Use the LINEAR equation for \(y\): \(x = 2 \Rightarrow y = 5\); \(x = -1 \Rightarrow y = 2\).