A19 - Simultaneous linear equations | AQA GCSE Mathematics

Two unknowns usually need two pieces of information. You will solve pairs of linear equations by elimination or substitution, and read approximate solutions from intersecting graphs. Each answer is a pair of values that must satisfy both equations at the same time.

Eliminating matching terms

Two numbers have sum $11$ and difference $3$ . What are they? Write $x+y=11$ and $x-y=3$ . One equation leaves many possible pairs; using both will pin down the pair.

You already know

You can solve a linear equation by doing the same operation to both sides. Collect like terms, and keep a negative sign with the term it belongs to. A point $(x,y)$ on a line satisfies that line's equation.

For $x+y=11$ , the pairs $(1,10)$ , $(2,9)$ and $(7,4)$ all work. Only the last also has difference $3$ . The second equation filters out the other possibilities.

Simultaneous solution

A pair of values that satisfies every equation in a given set at the same time. For example, $x=7$ , $y=4$ satisfies both $x+y=11$ and $x-y=3$ . The pair $x=6$ , $y=5$ satisfies the first equation only, so is not a simultaneous solution.

Elimination means adding or subtracting complete equations so that one unknown cancels. Equal terms cancel when subtracted; opposite terms cancel when added. Adding equal quantities on each side preserves equality. The elimination method is not on the formulae sheet: learn it.

Worked example 1: opposite terms cancel when added

Question. Solve $x+y=11$ and $x-y=3$ . Non-calculator.

Plan. Add the equations to eliminate $y$ , then substitute $x$ into either original equation.

\begin{aligned} (x+y)+(x-y)&=11+3&&\text{add whole equations}\\ 2x&=14&&y-y=0\\ x&=7&&\text{divide by }2\\ y&=11-7=4&&\text{substitute into }x+y=11 \end{aligned}

Why add? The coefficients of $y$ are $1$ and $-1$ , so their sum is zero.

Answer. $x=7$ , $y=4$ .

Check. $7+4=11$ and $7-4=3$ . Both equations work.

Change only the sign of the second right side. The same elimination steps now swap the two values.

Worked example 2: a negative difference

Question. Solve $x+y=11$ and $x-y=-3$ . Non-calculator.

Plan. Add to cancel $y$ , then substitute the new $x$ .

\begin{aligned} 2x&=11+(-3)=8&&\text{add the equations}\\ x&=4&&\text{divide by }2\\ y&=11-4=7&&\text{substitute} \end{aligned}

Why keep the negative on $-3$ ? It is the whole right side of the second equation and must be added as written.

Answer. $x=4$ , $y=7$ .

Check. $4+7=11$ and $4-7=-3$ .

Subtraction works just as well with equal terms. For $2x+5y=19$ and $2x+y=7$ , subtracting the second equation from the first gives $4y=12$ . You must subtract both the left and right sides.

Watch out

Wrong: adding the left sides but subtracting the right sides. Right: use the same operation on both sides. For $x+y=11$ and $x-y=3$ , adding gives $2x=14$ , not $2x=8$ .

Change the sum while keeping a negative difference.

Making coefficients match

When neither unknown cancels immediately, multiply one or both equations to create equal coefficients.

For example, doubling $x+y=7$ gives $2x+2y=14$ . Every solution of the original equation still satisfies this doubled equation. Multiply every term, including the right side. These equivalent-equation rules are not on the formulae sheet: learn them.

Worked example 3: multiply one equation

Question. Solve $2x+3y=17$ and $x+y=7$ . Non-calculator.

Plan. Double the second equation, subtract to eliminate $x$ , then substitute.

\begin{aligned} 2x+2y&=14&&\text{double the second equation}\\ (2x+3y)-(2x+2y)&=17-14&&\text{subtract}\\ y&=3&&\text{collect}\\ x&=7-3=4&&\text{use the second original equation} \end{aligned}

Why match $x$ ? One multiplication creates equal $2x$ terms, so subtraction removes them.

Answer. $x=4$ , $y=3$ .

Check. $2(4)+3(3)=17$ and $4+3=7$ .

Change only the second equation to $3x+2y=18$ . Now neither coefficient divides its partner, so multiply both equations.

Worked example 4: multiply both equations

Question. Solve $2x+3y=17$ and $3x+2y=18$ . Non-calculator.

Plan. Make both $x$ -coefficients $6$ , then subtract.

\begin{aligned} 6x+9y&=51&&\text{first equation times }3\\ 6x+4y&=36&&\text{second equation times }2\\ 5y&=15&&\text{subtract the second from the first}\\ y&=3&&\text{divide by }5\\ 2x+9&=17&&\text{substitute into the first}\\ x&=4&&\text{subtract }9\text{ then halve} \end{aligned}

Why choose $6x$ ? The least common multiple of $2$ and $3$ makes small matching coefficients. Matching $y$ instead is equally valid.

Answer. $x=4$ , $y=3$ .

Check. $2(4)+3(3)=17$ and $3(4)+2(3)=18$ .

Clear labels help keep the two scaled equations separate. A correct elimination can still be spoiled by substituting into a line copied with the wrong right side, so use an original equation.

Watch out

Wrong: doubling $x+y=7$ gives $2x+y=14$ . The $y$ term was not doubled. Right: $2x+2y=14$ .

Try the same coefficient pattern with different right sides.

Substituting one expression into another

If one equation already gives $y$ in terms of $x$ , replace $y$ in the other equation by that entire expression.

For $y=2x-1$ , every occurrence of $y$ can be replaced by $2x-1$ . This creates one equation in one unknown. Brackets protect the replacement when it is multiplied or subtracted. Substitution is another algebraic method; it is not on the formulae sheet.

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A19 - Simultaneous linear equations

Eliminating matching terms

Making coefficients match

Substituting one expression into another

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