2.5 Simultaneous equations by elimination and substitution — Learn

Prerequisites: solving a linear equation; expanding and factorising from 2.3; solving quadratics from 2.4; plotting a straight line and a curve.

By the end of this section you can: solve two linear equations by elimination; solve one linear and one quadratic equation by substitution; and interpret the solution as the point or points where the graphs meet.

1. Two equations at once

Two equations are simultaneous when they must be true at the same time. A solution is a pair of values $(x, y)$ that satisfies both equations. On a graph, each equation is a curve or a line, and each solution is a point where the two meet.

A line and a parabola can meet twice, touch once, or miss each other. This is the same discriminant idea as 2.3, seen on a graph.

2. Elimination

Use elimination when both equations are linear. Scale one or both equations so that one variable has the same coefficient in each, then add or subtract to remove it.

Example 1. Solve $3x + 2y = 16$ and $2x - y = 6$.

The second equation has a single $y$, so multiply it by $2$ to match the $2y$ in the first:

Now the $y$ terms have the same size and opposite signs. Add the equations to remove $y$:

So $x = 4$. Substitute into $2x - y = 6$:

The solution is $(4, 2)$. Check it in the other equation: $3 \times 4 + 2 \times 2 = 16$.

3. Substitution

Use substitution when one equation is linear and the other is quadratic. Rearrange the linear equation to make one variable the subject, substitute it into the quadratic, and solve the resulting quadratic.

Example 2. Solve $y = x + 1$ and $x^2 + y^2 = 25$.

The first equation already gives $y$ in terms of $x$. Substitute $x + 1$ for $y$:

Expand the bracket:

Collect terms and move $25$ to the left:

Divide by $2$, then factorise:

So $x = 3$ or $x = -4$. Find $y$ from $y = x + 1$ each time:

The two solutions are $(3, 4)$ and $(-4, -3)$: the points where the line meets the circle.

4. A line and a parabola

Example 3. Solve $y = x^2$ and $y = 2x + 3$.

Both expressions equal $y$, so set them equal to each other:

Rearrange to a quadratic in standard form:

Factorise:

so $x = 3$ or $x = -1$. Use $y = x^2$ to find each $y$:

The solutions are $(3, 9)$ and $(-1, 1)$.

5. A symmetric pair

Example 4. Solve $x + y = 5$ and $xy = 6$.

Rearrange the linear equation to make $y$ the subject:

Substitute into the second equation:

Expand and rearrange:

So $x = 2$ or $x = 3$. From $y = 5 - x$, the pairs are $(2, 3)$ and $(3, 2)$. The two answers are the same pair written in the two orders, because both equations treat $x$ and $y$ in the same way.

Watch out. Three slips cost marks here.

A sign error when subtracting. Multiplying an equation by a negative number changes every term. Write the scaled equation on its own line and check it before you add or subtract.

Stopping at $x$. A solution to two equations is a pair, so a value of $x$ is not an answer on its own. Find the value of $y$ for each $x$.

Dividing by a variable. Dividing by $x$ can remove the solution $x = 0$. Factorise instead. Re-check: substitute both coordinates into both of the original equations, not into the rearranged one.

Progress check

A straight service pipe runs through the tunnel along the line $y = \frac{x}{2} + 1$. The arch is $y = 4 - \frac{x^2}{4}$.

1. Write the two equations as a system and substitute the first into the second.
2. Solve the resulting quadratic.
3. The tunnel exists only for $-4 \le x \le 4$. Decide which solution lies inside the tunnel, and explain what happens to the other one.
4. Solve $x + y = 3$ and $xy = 2$ to check the method from Example 4.

You should now be able to

• decide whether elimination or substitution fits a pair of equations;
• eliminate a variable by scaling and adding or subtracting;
• substitute a linear expression into a quadratic equation;
• solve the resulting quadratic and find both coordinates;
• interpret two solutions as two points where the graphs meet.