2.15 - Solving simultaneous equations

Work out values that satisfy two equations at the same time, using algebra and graph intersections. You will solve two linear equations and systems containing a squared variable or a product, keeping the correct coordinates together in every solution.

Two linear equations together

Two numbers add to $7$ and multiply to $10$ . Does that determine just one ordered pair $(x,y)$ ? By the end of the lesson you will see why two pairs can work, and why a graph records both of them.

First consider linear equations. A single equation such as $x-2y=0$ allows many pairs. A second equation can narrow those possibilities to one pair that works in both.

Simultaneous solution

A solution of simultaneous equations in $x$ and $y$ is an ordered pair $(x,y)$ that satisfies every equation in the system. For example, $(4,2)$ satisfies both $x-2y=0$ and $3x+2y=16$ . A pair satisfying only one equation is not a simultaneous solution.

Elimination combines equations to remove one unknown. Multiply an entire equation when necessary, then add equations with opposite coefficients or subtract equations with equal coefficients. Afterwards, substitute the value you have obtained to recover the other unknown. These procedures are not on the formulae sheet: learn them.

Worked example 1: opposite coefficients

Question. Solve the simultaneous equations $3x+2y=16$ and $x-2y=0$ .

Plan. Add the equations to eliminate $y$ , then substitute for $x$ .

\begin{aligned} (3x+2y)+(x-2y)&=16+0 && \text{add the equations} \\ 4x&=16 && \text{the }y\text{-terms cancel} \\ x&=4 && \text{divide by }4 \\ 4-2y&=0 && \text{substitute into the second equation} \\ y&=2 && \text{solve for the remaining unknown} \end{aligned}

Why this step? The coefficients $+2$ and $-2$ cancel when added, leaving a one-unknown equation.

Answer. $x=4$ , $y=2$ , or $(x,y)=(4,2)$ .

Check. $3(4)+2(2)=16$ and $4-2(2)=0$ .

Worked example 2 changes only the coefficient of $x$ in the second equation. The same elimination works, but the values need not be integers.

Worked example 2: fractional values

Question. Solve $3x+2y=16$ and $2x-2y=0$ .

Plan. Eliminate $y$ by addition, then use the second equation to obtain $y$ .

\begin{aligned} 5x&=16 && \text{add the two equations} \\ x&=\frac{16}{5} && \text{divide by }5 \\ 2\left(\frac{16}{5}\right)-2y&=0 && \text{substitute} \\ y&=\frac{16}{5} && \text{solve the linear equation} \end{aligned}

Why this step? Equal opposite $y$ -coefficients still cancel; the changed $x$ -coefficient changes only the surviving equation.

Answer. $(x,y)=\left(\frac{16}{5},\frac{16}{5}\right)$ .

Check. $3\left(\frac{16}{5}\right)+2\left(\frac{16}{5}\right)=16$ , and the second equation gives zero.

Substitution is another option: express one unknown in terms of the other, then replace it in the other equation. In the first example, $x=2y$ gives $3(2y)+2y=16$ , so $y=2$ and $x=4$ . This is particularly useful when a system contains a quadratic, because elimination will not usually cancel the squared term.

Watch out

Wrong: multiply $x+2y=7$ by $3$ to get $3x+2y=21$ . Right: $3x+6y=21$ . Every term, including the constant, must be multiplied.

Try elimination with a different pair of opposite coefficients, keeping the fractional answer exact.

A line and a quadratic

When both equations give an expression for $y$ , equate those expressions to produce a quadratic in $x$ , then calculate a matching $y$ for every root.

You already know

Solve a quadratic by factorisation, completing the square or the formula. It may have two distinct roots, one repeated root or no real roots. Factorisation and completing-square rules are not on the formulae sheet; the quadratic formula is on it.

The two original equations still matter after substitution. The quadratic gives possible $x$ -values, but a solution of the system needs the corresponding $y$ -value too.

Worked example 3: two paired solutions

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

Plan. Equate the expressions for $y$ , solve for $x$ , then use the line to pair each root with $y$ .

\begin{aligned} x^2&=2x+3 && \text{both expressions equal }y \\ x^2-2x-3&=0 && \text{rearrange to zero} \\ (x-3)(x+1)&=0 && \text{factorise} \\ x&=3\text{ or }-1 && \text{solve each factor} \\ x=3&\quad\Longrightarrow\quad y=9 && \text{substitute in the line} \\ x=-1&\quad\Longrightarrow\quad y=1 && \text{substitute again} \end{aligned}

Why this step? The second value of $y$ must come from the second root, so pairing is part of the solution, not an optional final label.

Answer. $(x,y)=(3,9)$ or $(-1,1)$ .

Check. $9=3^2=2(3)+3$ , and $1=(-1)^2=2(-1)+3$ .

Worked example 4 changes only the line's constant term, from $3$ to $-1$ . The quadratic now has a repeated root, so there is only one distinct pair.

Worked example 4: one repeated root

Question. Solve $y=x^2$ and $y=2x-1$ .

Plan. Equate the two rules for $y$ , then check how many distinct roots result.

\begin{aligned} x^2&=2x-1 && \text{equate the outputs} \\ x^2-2x+1&=0 && \text{rearrange} \\ (x-1)^2&=0 && \text{a repeated factor} \\ x&=1 && \text{one distinct root} \\ y&=2(1)-1=1 && \text{recover the matching output} \end{aligned}

Why this step? A repeated root does not create a second point; both copies give the same $(x,y)$ .

Answer. $(x,y)=(1,1)$ .

Check. Both equations give $y=1$ when $x=1$ .

A graph will explain this change later: a line may cross a parabola twice, touch it once or miss it entirely. Algebra records these possibilities through the resulting quadratic.

Watch out

Wrong: report $x=3,-1$ and $y=9,1$ without showing which values go together. Right: write $(3,9)$ and $(-1,1)$ . The crossed pair $(3,1)$ satisfies neither equation here.

The next line has a new constant. Recover a second coordinate for each root before you stop.

Choosing the squared variable

If the second-order equation contains $y^2$ , it is often simplest to rearrange the line for $x$ , leaving a quadratic in $y$ .

For the line $y=x+2$ , write $x=y-2$ . Substituting this into an equation such as $y^2=3x+4$ keeps just one squared variable and avoids expanding a squared linear bracket. Both choices can work; choose the one with simpler algebra. This choice of substitution variable is not on the formulae sheet: learn it.

Worked example 5: a quadratic in y

Question. Solve $y=x+2$ and $y^2=3x+4$ .

Plan. Write $x=y-2$ , substitute into the second equation and solve for $y$ first.

\begin{aligned} y^2&=3(y-2)+4 && \text{substitute for }x \\ y^2-3y+2&=0 && \text{expand and rearrange} \\ (y-1)(y-2)&=0 && \text{factorise} \\ y&=1\text{ or }2 && \text{solve the quadratic} \\ y=1&\quad\Longrightarrow\quad x=-1 && x=y-2 \\ y=2&\quad\Longrightarrow\quad x=0 && x=y-2 \end{aligned}

Why this step? The variable you solve first need not be $x$ . Keeping the squared variable as the unknown makes the calculation shorter.

Answer. In $(x,y)$ order, the solutions are $(-1,1)$ and $(0,2)$ .

Check. $1^2=3(-1)+4=1$ , and $2^2=3(0)+4=4$ ; each pair also satisfies the line.

Worked example 6 changes only the constant $4$ to $10$ . One $y$ -value is now negative, which is allowed even though $y^2$ is non-negative.

Worked example 6: a negative y-coordinate

Question. Solve $y=x+2$ and $y^2=3x+10$ .

Plan. Use the same rearranged line and keep both roots of the quadratic in $y$ .

\begin{aligned} y^2&=3(y-2)+10 && \text{substitute for }x \\ y^2-3y-4&=0 && \text{collect terms} \\ (y-4)(y+1)&=0 && \text{factorise} \\ y&=4\text{ or }-1 && \text{retain both signs} \\ y=4&\quad\Longrightarrow\quad x=2 && x=y-2 \\ y=-1&\quad\Longrightarrow\quad x=-3 && x=y-2 \end{aligned}

Why this step? The equation contains $y^2$ , not a requirement that $y$ itself be positive.

Answer. $(x,y)=(2,4)$ or $(-3,-1)$ .

Check. $4^2=3(2)+10=16$ , and $(-1)^2=3(-3)+10=1$ . The line gives the stated $y$ -value in each case.

Do not take a square root too early and keep only its positive value. Solving the quadratic directly retains every possible sign. Also keep the coordinate order $(x,y)$ , even when $y$ was calculated first.

Watch out

Wrong: $y^2=1$ means only $y=1$ . Right: $y=1$ or $y=-1$ . A negative coordinate is allowed unless the original context or domain rules it out.

Change the constant again and carry both possible $y$ -values back to the line.

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2.15 - Solving simultaneous equations

Two linear equations together

A line and a quadratic

Choosing the squared variable

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