Simultaneous Equations GCSE Questions, Worked Examples and Answers

Substitute the pair into the first equation.

$3x+y=3(2)+3$

$3x+y=6+3=9\quad\checkmark$

Now check the second equation. A simultaneous solution must make this equation true as well.

$x+2y=2+2(3)$

$x+2y=2+6=8\quad\checkmark$

Both left-hand sides equal their required right-hand sides, so $\boxed{x=2,\ y=3}$ is a solution of the pair.

The $y$ coefficients are opposites, $+3$ and $-3$, so add the equations. This makes $3y+(-3y)=0$.

$\begin{aligned}2x+3y&=12\\4x-3y&=6\\\hline6x&=18\end{aligned}$

$x=3$

Substitute $x=3$ into the first original equation.

$2(3)+3y=12$

$6+3y=12$

$3y=6$

$y=2$

Check in the second equation: $4(3)-3(2)=12-6=6$. Therefore $\boxed{x=3,\ y=2}$.

The $y$ terms have the same sign and coefficient, so subtract the second equation from the first. Writing brackets helps the signs stay clear.

$\begin{aligned}(5x+2y)-(3x+2y)&=24-16\\2x&=8\end{aligned}$

$x=4$

Substitute $x=4$ into $3x+2y=16$.

$3(4)+2y=16$

$12+2y=16$

$2y=4$

$y=2$

Check in the other equation: $5(4)+2(2)=20+4=24$. Therefore $\boxed{x=4,\ y=2}$.

The $y$ terms are opposites, so add the equations.

$\begin{aligned}4x-y&=14\\2x+y&=4\\\hline6x&=18\end{aligned}$

$x=3$

Substitute into $2x+y=4$.

$2(3)+y=4$

$6+y=4$

Subtract $6$ from both sides.

$y=-2$

Check in the first equation: $4(3)-(-2)=12+2=14$. Therefore $\boxed{x=3,\ y=-2}$.

The $x$ coefficient $3$ is a multiple of $1$, so multiply the whole second equation by $3$.

$3(x+3y)=3(17)$

$3x+9y=51$

The $x$ terms now match with the same sign. Subtract the original first equation from the new equation.

$\begin{aligned}(3x+9y)-(3x+2y)&=51-16\\7y&=35\end{aligned}$

$y=5$

Substitute into $x+3y=17$.

$x+3(5)=17$

$x=2$

Check in the first equation: $3(2)+2(5)=6+10=16$. Therefore $\boxed{x=2,\ y=5}$.

Neither pair of coefficients matches. For $y$, the lowest useful common multiple of $4$ and $3$ is $12$.

Multiply every term in the first equation by $3$.

$9x+12y=54$

Multiply every term in the second equation by $4$.

$20x-12y=4$

The $y$ terms are now opposites, so add.

$29x=58$

$x=2$

Substitute into $3x+4y=18$.

$3(2)+4y=18$

$4y=12$

$y=3$

Check in the second equation: $5(2)-3(3)=10-9=1$. Therefore $\boxed{x=2,\ y=3}$.

The first equation already tells us what $y$ equals, so replace $y$ in the second equation with $4x-9$.

$2x+(4x-9)=9$

$6x-9=9$

$6x=18$

$x=3$

Use $y=4x-9$ to find $y$.

$y=4(3)-9$

$y=3$

Check in the other equation: $2(3)+3=9$. Therefore $\boxed{x=3,\ y=3}$.

Replace $x$ in the second equation with the equal expression $y+4$. Keep it in brackets because the whole expression is multiplied by $3$.

$3(y+4)-2y=17$

$3y+12-2y=17$

$y+12=17$

$y=5$

Then $x=y+4=5+4=9$.

Check: $3(9)-2(5)=27-10=17$. Therefore $\boxed{x=9,\ y=5}$.

Let $p$ be the number of pens and $n$ be the number of notebooks.

The total number of items gives

$p+n=18$

The total cost gives

$2p+5n=66$

Multiply the whole first equation by $2$ so that the $p$ terms match.

$2p+2n=36$

Subtract this from the cost equation.

$\begin{aligned}(2p+5n)-(2p+2n)&=66-36\\3n&=30\\n&=10\end{aligned}$

Substitute into $p+n=18$.

$p+10=18$

$p=8$

Check both facts: $8+10=18$ and $2(8)+5(10)=16+50=66$. The stall sold $\boxed{8\text{ pens and }10\text{ notebooks}}$.

Let the length be $l$ cm and the width be $w$ cm.

Since $2l+2w=38$, divide the perimeter equation by $2$.

$l+w=19$

The comparison in the question gives

$l=w+3$

Substitute $w+3$ for $l$ in the first equation.

$w+3+w=19$

$2w+3=19$

$2w=16$

$w=8$

Then $l=w+3=11$.

Check: $2(11)+2(8)=22+16=38$, and $11$ is $3$ more than $8$. The dimensions are $\boxed{11\text{ cm by }8\text{ cm}}$.

The solution is the coordinate of the point where the two lines intersect.

The graph shows the intersection at $(3,5)$, so

$\boxed{x=3,\ y=5}$

Check algebraically: $2(3)-1=5$ and $-3+8=5$. Both equations give the same $y$-value.

Both lines have gradient $2$, so they are parallel. Their intercepts are different, so they are not the same line and never meet.

Algebra gives the same conclusion. If both right-hand sides were equal, then

$2x+1=2x-4$

Subtracting $2x$ from both sides gives

$1=-4$

This contradiction cannot be true. Therefore the pair has $\boxed{\text{no solution}}$.

Both expressions equal $y$, so substitute $x+6$ for $y$ in the quadratic equation.

$x+6=x^2$

Move every term to one side.

$x^2-x-6=0$

Factorise.

$(x-3)(x+2)=0$

Therefore

$x=3\quad\text{or}\quad x=-2$

Each $x$-value needs its own matching $y$-value.

For $x=3$, $y=3+6=9$.

For $x=-2$, $y=-2+6=4$.

Checks: $3^2=9$ and $(-2)^2=4$. The two solution pairs are $\boxed{(3,9)\text{ and }(-2,4)}$.

Make $y$ the subject of the linear equation.

$y=5-x$

Substitute the whole expression into the circle equation. The brackets matter because $y$ is squared.

$x^2+(5-x)^2=13$

Expand the square.

$x^2+25-10x+x^2=13$

$2x^2-10x+12=0$

Divide every term by $2$.

$x^2-5x+6=0$

Factorise.

$(x-2)(x-3)=0$

So $x=2$ or $x=3$.

Using $y=5-x$ gives $y=3$ when $x=2$, and $y=2$ when $x=3$.

Both pairs satisfy $x^2+y^2=13$: $2^2+3^2=13$ and $3^2+2^2=13$. The solutions are $\boxed{(2,3)\text{ and }(3,2)}$.