Simultaneous Equations GCSE Questions, Worked Examples and Answers

Put $x=2$ and $y=3$ into the first equation.

$$3x+y=3(2)+3$$ $$3x+y=6+3=9\quad\checkmark$$

Now put the same values into the second equation. The pair must work in both.

$$x+2y=2+2(3)$$ $$x+2y=2+6=8\quad\checkmark$$

Both calculations are correct, so $\boxed{x=2,\ y=3}$ is the solution.

The $y$-terms are $+3y$ and $-3y$. Add the equations because these terms cancel to zero.

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

Put $x=3$ into the first equation to find $y$.

$$2(3)+3y=12$$ $$6+3y=12$$ $$3y=6$$ $$y=2$$

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

Both equations have $+2y$. Subtract the second equation from the first so $2y-2y=0$. Brackets help you keep the signs correct.

$$\begin{aligned}(5x+2y)-(3x+2y)&=24-16\\2x&=8\end{aligned}$$ $$x=4$$

Put $x=4$ into $3x+2y=16$ to find $y$.

$$3(4)+2y=16$$ $$12+2y=16$$ $$2y=4$$ $$y=2$$

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.

$$\begin{aligned}4x-y&=14\\2x+y&=4\\\hline6x&=18\end{aligned}$$ $$x=3$$

Put $x=3$ 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 first equation has $3x$ and the second has $x$. Multiply the whole second equation by $3$ to make another $3x$.

$$3(x+3y)=3(17)$$ $$3x+9y=51$$

The equations now both have $+3x$. Subtract the first equation from the new equation so the x-terms cancel.

$$\begin{aligned}(3x+9y)-(3x+2y)&=51-16\\7y&=35\end{aligned}$$ $$y=5$$

Put $y=5$ into $x+3y=17$.

$$x+3(5)=17$$ $$x=2$$

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

The numbers in front of $y$ are $4$ and $-3$. They do not match. The first number in both the 4 and 3 times tables 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.

$$29x=58$$ $$x=2$$

Put $x=2$ 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 $\boxed{x=3,\ y=3}$.

The first equation says $x=y+4$. Replace $x$ in the second equation with $(y+4)$. Use brackets because all of $y+4$ 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 new equation from the cost equation. The p-terms cancel.

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

Put $n=10$ 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 answer is the point where the two lines cross.

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

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

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

Both lines go up at the same rate: their gradient is $2$. They cross the y-axis at different points, so they are parallel and never meet.

You can also see this with algebra. If the lines met, the two expressions for $y$ would be equal.

$$2x+1=2x-4$$

Subtracting $2x$ from both sides gives

$$1=-4$$

$1=-4$ is impossible. Therefore the pair has $\boxed{\text{no solution}}$.

Both equations tell us what $y$ equals. This means $2x+3$ and $x^2$ must be equal.

$$2x+3=x^2$$

Move every term to one side.

$$x^2-2x-3=0$$

Factorise.

$$(x-3)(x+1)=0$$

Therefore

$$x=3\quad\text{or}\quad x=-1$$

There are two x-values. Find the matching y-value for each one.

For $x=3$, $y=2(3)+3=9$.

For $x=-1$, $y=2(-1)+3=1$.

Checks: $3^2=9$ and $(-1)^2=1$. The two solution pairs are $\boxed{(3,9)\text{ and }(-1,1)}$.

Start by getting $y$ on its own in the first equation.

$$y=5-x$$

Replace $y$ in the second equation with $5-x$. Use brackets because the whole expression 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 $x=3$.

Check both pairs: $2^2+3^2=13$ and $3^2+2^2=13$. The solutions are $\boxed{(2,3)\text{ and }(3,2)}$.