Simultaneous Equations with a Sum of Squares

When the second equation is a sum of squares (as in $x^2 + y^2 = 13$), the method is unchanged: make a variable the subject of the linear equation and substitute. The care is all in expanding the squared bracket correctly.

Solve the simultaneous equations $x + y = 5 \qquad \text{and} \qquad x^2 + y^2 = 13.$ [5]

The working

Step 1, make $y$ the subject of the linear equation: $y = 5 - x \quad \text{(M1)}$

Step 2, substitute into the quadratic and expand $(5 - x)^2 = 25 - 10x + x^2$: $x^2 + (5 - x)^2 = 13 \;\Rightarrow\; x^2 + 25 - 10x + x^2 = 13 \quad \text{(M1)}$

Step 3, simplify to a quadratic $= 0$: $2x^2 - 10x + 12 = 0 \;\Rightarrow\; x^2 - 5x + 6 = 0 \quad \text{(A1)}$

Step 4, factorise and pair up: $(x - 2)(x - 3) = 0 \;\Rightarrow\; x = 2 \text{ or } x = 3$ $x = 2 \Rightarrow y = 3, \qquad x = 3 \Rightarrow y = 2 \quad \text{(A1, A1)}$

Solutions $(2, 3)$ and $(3, 2)$. The symmetry (swapping $x$ and $y$) is expected here, because both original equations are symmetric in $x$ and $y$, a neat self-check.

Where the marks are won and lost

• Expand $(5 - x)^2$ in full: $25 - 10x + x^2$. Writing $25 - x^2$ or $25 + x^2$ is the error that sinks this question.
• Dividing $2x^2 - 10x + 12 = 0$ by $2$ gives the clean $x^2 - 5x + 6 = 0$.
• Pair each $x$ with the correct $y$ via $y = 5 - x$, not by guessing.

Common mistakes

• Expanding $(5 - x)^2$ incorrectly.
• Forgetting the second $x^2$ (from the substituted $y^2$), which changes the coefficient.
• Giving $x$-values only, or mispairing the coordinates.

Full method: Points of Intersection notes. Topic home: Simultaneous Equations pillar.