Solving simultaneous equations by substitution

To solve $2x + y = 7$ and $y = x^2 - x - 1$ simultaneously by substitution, what should you do first?

• Rearrange the linear equation to $y = 7 - 2x$, then substitute into the quadratic.

  Correct. The linear equation rearranges with no extra technique at all, and the quadratic is already isolated for y — substituting one directly into the other needs nothing further.

• BRearrange the quadratic by swapping x and y to get $x = y^2 - y - 1$, then substitute into the linear equation.

  Swapping the letters in an equation does not correctly invert the relationship between them — $x=y^2-y-1$ does not describe the same curve as $y=x^2-x-1$ solved the other way round. Relabelling is not the same operation as solving.

• CSubstitute $y=x^2-x-1$ into $2x+y=7$, giving $x^2+x-1=7$, and factorise the left-hand side directly to solve it.

  The substitution itself is correct ($2x+x^2-x-1=x^2+x-1$), but $x^2+x-1=7$ is not yet ready to factorise and solve — none of the three standard methods (factorising, the formula, completing the square) applies until the equation is rearranged into $=0$ form: $x^2+x-8=0$.

• DSolve $y=x^2-x-1$ for x using the quadratic formula, then substitute that expression for x into $2x+y=7$.

  This can technically be forced through, but it needlessly introduces a square root into x's expression, for no reason: the linear equation was already the simple one to rearrange, and the quadratic was already isolated for y. Never rearrange the quadratic when the linear equation is available to rearrange instead.

Traps tested: Variables swapped instead of solved for · Quadratic not rearranged to equal zero before solving · Quadratic rearranged instead of the linear equation

After substituting a line into a quadratic curve, you correctly reach $x^2+3x+10=0$. What does this tell you about the line and the curve?

• They never meet — $b^2-4ac=9-40=-31$, and a negative discriminant means the equation has no real solutions.

  Correct. $3^2-4(1)(10)=9-40=-31<0$. No real x satisfies the equation, so there is no point common to both the line and the curve.

• BThey meet twice, since every quadratic equation has exactly two roots.

  A quadratic equation has two REAL roots only when its discriminant is positive. Here $b^2-4ac=-31<0$, so it has no real roots at all — 'quadratic' describes the shape of the equation, not a guarantee of two real solutions.

• CThey are tangent — a negative discriminant is the boundary case where the line just touches the curve.

  The boundary case for a tangent is $b^2-4ac=0$ exactly. A negative discriminant is on the far side of that boundary from a positive one, in the direction of no real intersection at all, not the direction of a single touching point.

• DIt cannot be determined without also knowing the y-values of the intersection points.

  The discriminant of the equation substitution produces settles the question completely on its own — a negative value means there is no real x, so there is no intersection point to find a y-value for.

Traps tested: Assumes every quadratic has two real roots · Negative discriminant read as tangency · Overclaims uncertainty about real root count

Solving $y=x-1$ and $y=x^2-3x-1$ simultaneously, you correctly reach $x^2-4x=0$ and solve it to get $x=0$ or $x=4$. What is the correct final answer?

• $(0,-1)$ and $(4,3)$ — found by substituting each x-value back into $y=x-1$.

  Correct. At $x=0$: $y=0-1=-1$. At $x=4$: $y=4-1=3$. Both pairs check against the curve too: $0-0-1=-1$ ✓ and $16-12-1=3$ ✓.

• B$x=0$ or $x=4$

  This states the x-coordinates, not the points of intersection. Each value of x needs converting back into a full coordinate pair — this response stops one step short of what the question asked for.

• C$(0,0)$ and $(4,4)$ — read the coordinates straight from the two values of x.

  This assumes y always equals x, which is only true for a point that lies on the line $y=x$ — not the line in this question. The actual y-value has to be calculated by substitution, not assumed from the x-value alone.

• DOnly $(4,3)$, since $x=0$ gives $y=-1$, and coordinates cannot be negative.

  Coordinates can absolutely be negative — nothing in either original equation restricts x or y to be positive. This is unlike the hidden-quadratic-substitution lesson's own domain restriction, where a negative value of the substituted letter really does get rejected because it can never equal $a^x$ for $a>0$; there is no such restriction here, and $(0,-1)$ is a completely genuine solution.

Traps tested: Stops at the x coordinate · Y value assumed equal to x value · False belief that coordinates cannot be negative

Three vertices of a rectangle are given, and you need to find the fourth. Which statement is correct?

• Solving simultaneously for the equations of the two unknown sides is the method spec 1.6 names directly, and it always works — but a real examiner report on this exact question type records it scoring less reliably in practice than the diagonal-midpoint shortcut, because there is more that has to go right along the way.

  Correct — and this is exactly the finding this lesson's method-comparison block is built around. Both methods are mathematically valid; the practical difference is how much correctly-executed algebra has to survive before an answer appears.

• BThe diagonal-midpoint method isn't creditable, since it isn't 'solving simultaneous equations' in the way spec 1.6 describes.

  A mark scheme credits any valid method reaching the correct answer — the same 'any valid method' principle that already applies to solving a three-term quadratic. The diagonal-midpoint approach is every bit as valid a piece of mathematics; it's simply built on a different fact (parallelogram diagonals bisect each other) rather than on gradients and line equations.

• CThe simultaneous-equations method should never be used for this type of question, since a shortcut always exists.

  Not every version of this question hands you a shape with an available shortcut — three points forming an ordinary triangle, say, rather than three vertices of a special quadrilateral. The mechanical simultaneous-equations method is the one that always works, whatever the shape; a shortcut is a bonus when the geometry happens to offer one, not a replacement for knowing the general method.

• DBoth methods are equally reliable in practice, so the choice makes no real difference to the mark you're likely to score.

  This contradicts the real, quoted finding directly: candidates using the simultaneous-equations route 'rarely produced enough work to score full marks,' while the diagonal-midpoint route was 'often successful' (Jan 2023, Q2). The two routes are not equally reliable in practice, even though both are mathematically valid.

Traps tested: Shortcut method assumed uncreditable · Overgeneralizes that a shortcut always exists · Assumes equal reliability despite quoted finding

Two different WMA11 topics both use the word 'substitution.' Which statement correctly distinguishes them?

• Spec 1.6's substitution eliminates a second EQUATION, using one to remove a variable from another; the hidden-quadratic technique eliminates an unfamiliar TYPE of expression (like $3^x$) within a single equation, using a new letter instead.

  Correct — the two techniques share only their name and their closing discipline (solve the equation produced, then convert back to what was actually asked for). One collapses two equations into one; the other collapses one unfamiliar equation into a familiar one.

• BThey are the same technique, just applied to different-looking equations — there's no real difference.

  This flattens a genuine structural difference: spec 1.6's version needs TWO equations in the first place, and produces a genuinely new variable count reduction; the hidden-quadratic technique works on a single equation, and its new letter u still refers to the exact same quantity as before ($u=3^x$), never a second unknown eliminated by a second equation.

• CSpec 1.6's substitution only works when both equations are linear; the hidden-quadratic technique is for anything involving a quadratic.

  Spec 1.6's substitution is exactly for pairing a linear equation with a quadratic curve, among other combinations — it is not restricted to two linear equations, even though one worked example in this lesson happens to use two.

• DThe hidden-quadratic technique is part of spec 1.6, since both are called 'substitution.'

  The hidden-quadratic-substitution lesson is scoped to spec 1.1 (laws of indices) plus 1.5 (solving the resulting quadratic) — not spec 1.6. Sharing a name doesn't put two techniques under the same spec point; the actual mechanism (one equation vs two) is what settles that.

Traps tested: Distinct substitution techniques conflated · Spec 1 6 substitution wrongly restricted to linear pairs · Topics conflated by shared vocabulary