A-Level Simultaneous equations Practice Questions

Label the two equations

We label them (1) and (2) so we can refer to each equation clearly. The plan is to eliminate one letter so we are left with a single equation in one unknown.

Add the equations

Because the $y$-terms are equal and opposite we ADD, which cancels $y$. This leaves one equation with only $x$ in it.

Solve for $x$

Dividing both sides by the number in front of $x$ isolates $x$. This is the same rearranging skill used all through algebra.

Substitute back to find $y$

Put the value of $x$ into equation (1) and solve the small equation that remains. Using the original equation keeps the numbers exact.

State the solution

This ordered pair is where the two straight lines cross on a graph, so it is the single point that satisfies both equations at once.

Simplify the second equation

Dividing the whole second equation by 2 puts it in the same form as the first. Dividing every term keeps it balanced.

Compare the two equations

Both equations are now identical, so they describe exactly the same straight line.

Interpret one line drawn twice

If both equations are the same line, then every single point on that line satisfies both — there is no unique crossing point.

Rearrange each equation to $y = mx + c$

Putting both lines in gradient-intercept form lets us picture their slopes and where they cut the $y$-axis.

Solve the system algebraically

Solving simultaneously tells us exactly where the correct graph should show the lines crossing.

Locate the intersection on the graph

The right diagram is the one whose two lines have the correct gradients and intercepts and cross at this point.

Reject the distractors

The other diagrams either mark the wrong crossing point, show parallel lines (no solution), or have a line reflected or shifted, so they do not match both equations.

Confirm the choice

Both lines pass through the marked point, confirming the selected diagram is correct.

Note the gradients and intercepts

The correct graph must also show these gradients and $y$-intercepts, not just the crossing point — a useful way to eliminate distractors.

Set the two expressions for $y$ equal

Both equations give $y$, and at an intersection the $y$-values are equal. Setting them equal turns the pair into a single equation in $x$.

Move everything to one side

We collect all terms on one side so the equation equals zero. This is the standard form for solving a quadratic, just like earlier quadratic work.

Identify the coefficients

Reading off $a$, $b$ and $c$ prepares us to factorise or use the quadratic formula. Careful signs here avoid mistakes later.

Apply the quadratic formula

When the quadratic does not factorise nicely we use the formula. It always works, whatever the coefficients.

Simplify the discriminant

The number under the root is the discriminant. Because it is positive there are two separate intersection points.

Solve for $x$

Simplifying the surd (or fraction) gives the two exact $x$-coordinates of the intersections.

Find $y$ for the first point

Substitute the first $x$-value into the simpler (linear) equation to get its $y$-value. Always use the line — it is less work than the curve.

Find $y$ for the second point

Do the same with the second $x$-value. Now each $x$ has its matching $y$.

State both intersection points

These are the two points where the straight line crosses the curve on the graph.

Check the first point on the curve

Substituting back into the curve confirms the point really lies on it. A quick check catches arithmetic slips.

Equate line and curve

Intersection points satisfy both equations, so we set the $y$-expressions equal.

Form a quadratic in $x$

Rearranging gives a quadratic in $x$ whose real roots are the intersection points; its coefficients contain $k$.

Identify the coefficients

We list $a$, $b$, $c$ ready to build the discriminant.

Write the discriminant

For two distinct intersections we need the discriminant to be strictly positive.

Simplify the discriminant

Expanding gives a neat expression in $k$. We now show it is always positive.

Recognise a non-negative square

Writing it as a square plus a positive constant makes the sign obvious, because any real square is at least zero.

Deduce the minimum value

The smallest the expression can be is the positive constant, reached when the square is zero. So it is always positive.

Conclude two roots for all $k$

A strictly positive discriminant guarantees two distinct real roots, hence two intersection points, whatever the value of $k$.

Locate the vertex of the discriminant parabola

Viewing the discriminant as a parabola in $k$, its lowest point is where its rate of change is zero. This is where it is smallest.

Evaluate the minimum discriminant

Even at its smallest the discriminant is positive, so it is positive everywhere — the strongest form of the argument.

Interpret geometrically

Because the discriminant never reaches zero, the line is never a tangent and never misses — it always crosses the curve twice.

State the key idea

The whole proof rests on the fact that a real number squared cannot be negative, a technique used throughout A-Level inequalities and proof.

Confirm with a value

Checking a single value is consistent with the general proof and guards against slips.

Confirm with another value

A second check reinforces that the discriminant stays positive.

Write the conclusion carefully

A clear final sentence, quoting the discriminant being strictly positive, is what earns the proof marks.