Solving Equations and Simultaneous Equations

A linear equation has the variable to the first power only. Solve by performing inverse operations to isolate the variable.

Single-sided:

$3x+7=19⟹3x=12⟹x=4$

Unknown on both sides — collect variable terms on one side:

$5x-3=2x+9⟹3x=12⟹x=4$

With fractions — multiply through by the LCM of denominators:

$\frac{x+1}{3}=\frac{x-2}{2}⟹2(x+1)=3(x-2)⟹2x+2=3x-6⟹x=8$

With brackets — expand first:

$4(2x-1)=3(x+5)⟹8x-4=3x+15⟹5x=19⟹x=\frac{19}{5}$

A quadratic equation has the form $a{x}^2+bx+c=0$. The standard method at Foundation tier is factorising.

Method:

1. Rearrange so one side equals zero.
2. Factorise the left side.
3. Set each factor equal to zero.
4. Solve each resulting linear equation.

Worked example — solve $x^2-5x+6=0$:

$x^2-5x+6=(x-2)(x-3)=0$

$x=2\text{or}x=3$

Worked example — solve $2x^2+5x-3=0$ (Higher):

Factorise (AC method): $ac=-6$; two numbers multiplying to $-6$ and adding to $5$: $6$ and $-1$.

$2x^2+6x-x-3=2x(x+3)-1(x+3)=(2x-1)(x+3)=0$

$x=\frac{1}{2}\text{or}x=-3$

When a quadratic does not factorise neatly, use the quadratic formula:

$x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}\text{for }a{x}^2+bx+c=0$

Worked example — solve $2x^2-3x-4=0$ (give answers to 2 d.p.):

$x=\frac{3\pm \sqrt{9+32}}{4}=\frac{3\pm \sqrt{41}}{4}$

$x=\frac{3+6.403}{4}\approx 2.35\text{or}x=\frac{3-6.403}{4}\approx -0.85$

The discriminant $b^2-4ac$ tells you how many real roots exist:

• $b^2-4ac>0$: two distinct real roots
• $b^2-4ac=0$: one repeated root
• $b^2-4ac<0$: no real roots

Solving by completing the square:

$x^2+6x-7=0⟹(x+3{)}^2-9-7=0⟹(x+3{)}^2=16⟹x+3=\pm 4⟹x=1\text{ or }x=-7$

When an equation cannot be solved exactly by inspection, or when the question asks for an estimate, graph intersection provides approximate solutions.

Method — solving $f(x)=g(x)$ graphically: plot both $y=f(x)$ and $y=g(x)$ on the same axes. The $x$-coordinates of the intersection points are the approximate solutions.

Worked example — linear equation from a graph (A17):

To solve $3x-2=x+4$, plot $y=3x-2$ and $y=x+4$. The lines cross at $(3,7)$, so $x=3$.

This matches the algebraic solution: $3x-2=x+4⟹2x=6⟹x=3$ ✓

Worked example — quadratic from a graph (A18):

To solve $x^2-x-3=0$ approximately, plot $y=x^2-x-3$. Read off where the graph crosses the $x$-axis (where $y=0$).

Reading from the graph: $x\approx -1.3$ or $x\approx 2.3$ (to 1 d.p.)

Alternatively, rearrange to $x^2=x+3$, plot $y=x^2$ and $y=x+3$, and read the $x$-coordinates of the intersections.

Worked example — simultaneous equations from a graph (A19):

To solve $y=2x-1$ and $y=x+3$ simultaneously, plot both lines. They intersect at approximately $(4,7)$, giving $x\approx 4$, $y\approx 7$.

Algebraic check: $2(4)-1=7$ ✓ and $4+3=7$ ✓. The intersection point is the exact solution.

Key exam skill: read $x$-values at intersections or axis crossings to 1 decimal place unless the question specifies otherwise. Show which graph you are reading from and label the intersection.

Two simultaneous equations are solved when values of $x$ and $y$ satisfy both at once.

Elimination method — multiply equations to match coefficients, then add or subtract:

Worked example — solve $3x+2y=13$ and $5x-2y=11$:

Add: $8x=24⟹x=3$. Substitute: $9+2y=13⟹y=2$. Solution: $(3,2)$

Check in both: $3(3)+2(2)=13$ ✓; $5(3)-2(2)=11$ ✓

Substitution method:

Worked example — solve $y=3x-1$ and $2x+y=9$:

Substitute $y=3x-1$ into the second: $2x+(3x-1)=9⟹5x=10⟹x=2$

$y=3(2)-1=5$. Solution: $(2,5)$ ✓

When one equation is quadratic, use substitution to reduce to a single quadratic.

Worked example — solve $y=2x+1$ and $y=x^2+x-1$:

$2x+1=x^2+x-1⟹x^2-x-2=0⟹(x-2)(x+1)=0$

$x=2:y=5x=-1:y=-1$

Solutions: $(2,5)$ and $(-1,-1)$ ✓

Geometrically, these are the intersection points of a line and a parabola.

Iteration finds approximate solutions to equations numerically by repeatedly applying a rearrangement of the equation.

Method: rearrange the equation into the form $x=f(x)$. Starting from an initial estimate $x_0$, compute $x_1=f(x_0)$, then $x_2=f(x_1)$, and so on. The sequence converges to the root.

Worked example — show that $x^3-2x-5=0$ can be rearranged to $x=\sqrt[3]{2x+5}$, and use iteration with $x_0=2$ to find the root to 3 s.f.

Rearrangement: $x^3=2x+5⟹x=(2x+5{)}^{1/3}$ ✓

$x_1=(2(2)+5{)}^{1/3}=9^{1/3}\approx 2.0801$

$x_2=(2(2.0801)+5{)}^{1/3}=9.1602^{1/3}\approx 2.0924$

$x_3=(2(2.0924)+5{)}^{1/3}=9.1848^{1/3}\approx 2.0942$

$x_4=(2(2.0942)+5{)}^{1/3}=9.1884^{1/3}\approx 2.0945$ — converging to $x\approx 2.09$ (3 s.f.) ✓

1. Not rearranging to zero before factorising a quadratic

$x^2=3x+4$ must be rearranged to $x^2-3x-4=0$ before factorising. Factorising only works when one side equals zero.

2. Simultaneous equations — arithmetic error after elimination

After finding one variable, substitute back into one of the original equations (not a derived equation) and check the answer in the other equation.

3. Quadratic formula — sign error with $\pm$

$\pm \sqrt{b^2-4ac}$ gives two solutions. Writing only the $+$ case means the second solution is missed — common when a question asks for two answers.

4. Linear-quadratic — expecting only one solution

A line can intersect a parabola at 0, 1, or 2 points. Unless the discriminant is zero or the equations produce an impossible result, there are usually two solution pairs to find and state.

Mistake	Correction
"$x^2-5x=0$ has only solution $x=5$"	Factorise: $x(x-5)=0$; solutions $x=0$ and $x=5$ — don't divide both sides by $x$
"Solve $x^2=9$: $x=3$"	$x=\pm 3$ (both square roots: $3$ and $-3$)
"Iteration diverges, so the formula is wrong"	If iteration diverges, try a different rearrangement of the equation