Simultaneous Equations Worked ExamplesIGCSE Maths 0580

The $y$ coefficients are $+2$ and $-2$ — equal size, opposite signs — so adding eliminates $y$.

Simplify and solve for $x$.

Substitute back into either original equation.

Check in the other equation: $5(3) - 2(3.5) = 15 - 7 = 8$. ✓

No coefficients match, so scale both equations to make the $y$ terms equal in size. The LCM of $3$ and $2$ is $6$.

Scale the second equation.

The $y$ terms are now $+6y$ and $-6y$, so add.

Substitute into the first original equation.

Check in the second: $5(2) - 2(3) = 10 - 6 = 4$. ✓

The first equation already gives $y$ in terms of $x$, so substitution is quicker than elimination here.

Expand the bracket.

Collect and solve.

Substitute into the simpler equation.

Check: $3(2) + 2(3) = 12$. ✓

Define the variables explicitly, with units.

Write one equation per statement.

Double the first equation so the $p$ terms match.

Subtract the second equation from this — the signs match, so subtract.

Substitute back and check in the other equation.

Both equations give $y$, so set the right-hand sides equal. This eliminates $y$ in one step.

Rearrange so the quadratic equals zero.

Factorise — two numbers multiplying to $-4$ and adding to $-3$.

Find the matching $y$ for each $x$, using the linear equation — it is simpler and less error-prone.

Check in the quadratic: $4^2 - 2(4) = 8$ ✓ and $(-1)^2 - 2(-1) = 3$ ✓.

Clear the fractions first by multiplying the whole first equation by $6$.

Scale the second equation so the $y$ terms match in size.

Signs are opposite, so add.

Substitute into $x - y = 1$.

Check in the original first equation: $\frac{5.2}{2} + \frac{4.2}{3} = 2.6 + 1.4 = 4$. ✓

(a) Opposite sides of a rectangle are equal. The top equals the bottom, and the left side equals the right side.

Tidy the second equation by collecting the $x$ terms on one side and the $y$ terms on the other.

(b) Make the $y$ coefficients match: multiply the first equation by $4$, then add to eliminate $y$.

Substitute $x = 4$ into $2x + y = 11$. Check with the sides: left $= 4 + 6 = 10$ and right $= 16 - 6 = 10$ ✓.

(c) The rectangle measures $11\text{ cm}$ by $10\text{ cm}$.

(a) Read two points on $L$ where it crosses grid lines exactly: $(0, 3)$ and $(9, 0)$. The gradient is rise over run.

The $y$-intercept is $3$. Write $y = mx + c$, then multiply by $3$ to clear the fraction and rearrange into the requested form.

(b) The solution is where the lines cross. Reading from the small squares, the crossing point is at about $x = 1.7$, $y = 2.4$. Any reading within about one small square is acceptable for an estimate.

(c) Substitute $y = 2x - 1$ into $x + 3y = 9$.

Substitute back into $y = 2x - 1$. As decimals these are $1.71\ldots$ and $2.43\ldots$, which agree with the estimate from (b).

(a) A gradient of $2$ and $y$-intercept $-4$ give the equation of $L$.

At $A$ and $B$ the two $y$-values are equal, so set the expressions equal and rearrange to $= 0$. This is the given equation.

(b) Factorise: two numbers multiplying to $5$ and adding to $-6$.

Find each $y$ from the line (the simpler equation), and check in the curve. From the diagram, $A$ is the lower-left point.

Check: $1 - 4 + 1 = -2$ ✓ and $25 - 20 + 1 = 6$ ✓. So $A$ is $(1, -2)$ and $B$ is $(5, 6)$.

(c) Use Pythagoras on the horizontal and vertical changes from $A$ to $B$.

Simplify the surd using the largest square factor of $80$, which is $16$.

Both equations give $y$, so set them equal to eliminate $y$.

Collect all the terms on one side to make a three-term quadratic.

It does not factorise, so use the formula with $a = 1$, $b = -7$, $c = 3$. Write $(-7)^{2} = 49$ in your working — it is positive.

Evaluate both values of $x$, keeping the full calculator values.

Find each $y$ from the linear equation $y = 3x - 2$, using the unrounded $x$-values, then round both coordinates.

Opposite angles of a parallelogram are equal, so angle $K$ = angle $M$.

Angles $K$ and $L$ are co-interior angles between the parallel sides $KN$ and $LM$, so they add up to $180^{\circ}$.

The first equation already gives $b$ in terms of $a$, so substitute $b = 2a$ into the second.

Then use $b = 2a$.

Check with the angles: $K = 40 + 20 = 60^{\circ}$, $M = 20 + 40 = 60^{\circ}$ ✓ and $L = 10 + 100 + 10 = 120^{\circ}$, with $60 + 120 = 180$ ✓.