Substitution
Simultaneous equations are two equations with two unknowns that are true at the same time. In the substitution method, one equation is written with one pronumeral as the subject, then substituted into the other, as with \(y=2x-5\) and \(3x+y=15\). In NSW Year 10 Advanced maths the equations may contain brackets, fractions or decimals (NESA outcome MA5‑EQU‑P‑02).
You will solve pairs of linear equations by substitution, rearranging first when neither pronumeral is the subject, and simplify equations with brackets and fractions before solving. You will work out a line through two points, show that three lines meet at one point, and see what happens when the equations describe the same line or parallel lines. Two water tanks, one filling and one leaking, are compared. It is part of Year 10 Advanced Simultaneous Equations.
Every question with a fully worked solution.
- Substitution - Video - Solving simultaneous equations by substitution Watch
Theory
Two equations with two unknowns are simultaneous when the same \(x\) and \(y\) must make both true. In the substitution method you make one pronumeral the subject of one equation and put that expression into the other, leaving one equation with one unknown. This Year 10 Advanced maths page covers rearranging first, fractions and decimals, when there is no solution or infinitely many, and equations whose coefficients are unknown.
Simultaneous equations are two (or more) equations that are true at the same time. For two linear equations in \(x\) and \(y\), the solution is the pair \(x = \ldots,\ y = \ldots\) that satisfies both. On a graph it is the point where the two lines meet.
Substitution means replacing a pronumeral by an equal expression. If \(y = 2x - 5\), every \(y\) in the other equation can be replaced by \((2x - 5)\).
Substitution suits a pair where one equation already has \(y = \ldots\) or \(x = \ldots\), or where a pronumeral has coefficient \(1\) or \(-1\) and is easy to make the subject.
Two lines with the same gradient and different \(y\)-intercepts are parallel and have no solution. Two equations that rearrange to the same line have infinitely many solutions.
If one equation gives \(y\) in terms of \(x\), substitute it into the other:
If both equations give \(y\), set the two expressions equal:
The substitution method
- Tidy each equation first: expand brackets, collect like terms, and multiply out fractions or decimals.
- Label the equations (1) and (2).
- Make a pronumeral the subject of one equation, choosing one with coefficient \(1\) or \(-1\) if you can.
- Substitute that expression, in brackets, into the other equation, and solve for the one remaining unknown.
- Back-substitute into the rearranged equation to find the second unknown.
- Check both values in the equation you did not use, preferably the original form.
Other uses of the same idea
- Line through two points: put each point into \(y = mx + c\) to get two equations in \(m\) and \(c\), then solve them.
- Do three lines meet at one point? Solve two of them, then test the point in the third.
- A point with an unknown, such as \((t,\ 2t - 1)\) on two lines: substitute into the line with only one unknown first.
- Known solution, unknown coefficients: substitute the solution to get equations in the coefficients (Example 4).
- Word problems: write one equation per fact (for example \(V = 500 + 30t\) for a tank being filled), then substitute one into the other. Answer in words with units.
- After substituting: a statement such as \(0 = 5\) means no solution (parallel lines); \(4 = 4\) means infinitely many (the same line).
Label the equations (1) and (2). In (1) \(y\) has coefficient \(-1\), so it is the easiest pronumeral to make the subject.
| \(3x - y\) | \(=\) | \(4 \quad (1)\) |
| \(5x + 2y\) | \(=\) | \(25 \quad (2)\) |
| \(\text{From (1): } y\) | \(=\) | \(3x - 4 \quad (3)\) |
Substitute (3) into (2). Keep the brackets.
| \(5x + 2(3x - 4)\) | \(=\) | \(25\) |
| \(5x + 6x - 8\) | \(=\) | \(25\) |
| \(11x\) | \(=\) | \(33\) |
| \(x\) | \(=\) | \(3\) |
| \(\text{In (3): } y\) | \(=\) | \(3(3) - 4 = 5\) |
Check in (2), the equation not used to find \(y\): \(5(3) + 2(5) = 15 + 10 = 25\ \checkmark\)
| \(\therefore\ x\) | \(=\) | \(\boldsymbol{3},\quad y = \boldsymbol{5}\) |
Multiply every term of (1) by the lowest common denominator, \(12\). Each numerator goes into a bracket.
| \(4(x - 1) + 3(y + 2)\) | \(=\) | \(36\) |
| \(4x - 4 + 3y + 6\) | \(=\) | \(36\) |
| \(4x + 3y\) | \(=\) | \(34 \quad (3)\) |
Equation (2) already has \(x\) as the subject. Substitute it into (3).
| \(4(3y + 1) + 3y\) | \(=\) | \(34\) |
| \(12y + 4 + 3y\) | \(=\) | \(34\) |
| \(15y\) | \(=\) | \(30\) |
| \(y\) | \(=\) | \(2\) |
| \(x\) | \(=\) | \(3(2) + 1 = 7\) |
Check in the original (1): \(\dfrac{7 - 1}{3} + \dfrac{2 + 2}{4} = 2 + 1 = 3\ \checkmark\)
| \(\therefore\ x\) | \(=\) | \(\boldsymbol{7},\quad y = \boldsymbol{2}\) |
Decimals work the same way: multiply by \(10\) or \(100\) to clear them before substituting.
(i) Substitute \(y = kx + 4\) into \(3x - y = 2\) and show that \((3 - k)x = 6\).
(ii) For which value of \(k\) do the lines not meet?
(iii) Solve the pair when \(k = 5\).
(iv) What happens with \(y = 3x + 4\) and \(6x - 2y = -8\)?
(i) Replace \(y\) by \(kx + 4\) inside a bracket, because of the minus sign.
| \(3x - (kx + 4)\) | \(=\) | \(2\) |
| \(3x - kx\) | \(=\) | \(6\) |
| \((3 - k)x\) | \(=\) | \(6\) |
(ii) If \(k = 3\) the equation becomes \(0 \times x = 6\), which no \(x\) satisfies. Both lines then have gradient \(3\) with different \(y\)-intercepts, so they are parallel: \(\boldsymbol{k = 3}\).
(iii) With \(k = 5\):
| \(-2x\) | \(=\) | \(6\) |
| \(x\) | \(=\) | \(-3\) |
| \(y\) | \(=\) | \(5(-3) + 4 = -11\) |
| \(\therefore\ x\) | \(=\) | \(\boldsymbol{-3},\quad y = \boldsymbol{-11}\) |
(iv) Substitute:
| \(6x - 2(3x + 4)\) | \(=\) | \(-8\) |
| \(6x - 6x - 8\) | \(=\) | \(-8\) |
| \(-8\) | \(=\) | \(-8\) |
The \(x\) terms cancel and the statement is always true. The second equation is the first one doubled, so both give the same line: there are infinitely many solutions.
Substitute the known \(x\) and \(y\). That gives two equations in the new unknowns \(a\) and \(b\). Watch the signs: \(a(-3) = -3a\) and \(-b(2) = -2b\).
| \(-3a - 2b\) | \(=\) | \(13 \quad (1)\) |
| \(-3b + 2a\) | \(=\) | \(0 \quad (2)\) |
| \(\text{From (2): } a\) | \(=\) | \(\dfrac{3b}{2}\) |
Substitute into (1), then multiply by \(2\) to clear the fraction.
| \(-3 \times \dfrac{3b}{2} - 2b\) | \(=\) | \(13\) |
| \(-9b - 4b\) | \(=\) | \(26\) |
| \(-13b\) | \(=\) | \(26\) |
| \(b\) | \(=\) | \(-2\) |
| \(a\) | \(=\) | \(\dfrac{3(-2)}{2} = -3\) |
Check: \((-3)(-3) - (-2)(2) = 9 + 4 = 13\ \checkmark\) and \((-2)(-3) + (-3)(2) = 0\ \checkmark\)
| \(\therefore\ a\) | \(=\) | \(\boldsymbol{-3},\quad b = \boldsymbol{-2}\) |
Common pitfalls
Frequently asked questions
How do you solve simultaneous equations by substitution?
Make one pronumeral the subject of one equation, substitute that expression into the other equation, solve for the remaining pronumeral, then substitute back to find the first one and check in both equations.
When should I use substitution instead of elimination?
Substitution is quickest when one equation already reads y equals or x equals, or when a pronumeral has a coefficient of 1 or minus 1. Elimination is often easier when both equations are in the form ax plus by equals c.
What does it mean if the pronumerals cancel out?
If you are left with a false statement such as 0 equals 6, there is no solution and the lines are parallel. If you are left with a true statement such as 8 equals 8, the equations describe the same line and there are infinitely many solutions.
How do you find the equation of a line through two points with simultaneous equations?
Substitute each point into y equals mx plus c. This gives two equations in m and c. Solve them simultaneously, then write the line with those values.
How do I check my answer?
Substitute both values into each original equation. Both must be true. Checking in the equation you did not use to find the second value catches most errors.