Ma4, Ma5 - Simultaneous equations and intersections | OCR A-Level Mathematics B (MEI)

Use elimination and substitution for two linear equations in Ma4, then extend those tools to a linear and a quadratic equation in Ma5. Learn to recover the correct pairs of values and justify how many common points the equations describe.

Eliminating a variable

Could changing just one number turn a system with no solutions into one with infinitely many? Take $x+y=3$ and $2x+2y=7$ . The first equation requires $2x+2y=6$ , so the pair is impossible. Change $7$ to $6$ , though, and every point with $y=3-x$ works. The answer really can jump from none to infinitely many.

You already know

An equation can be multiplied by a non-zero constant without changing its solutions. Checking a pair means substituting both coordinates into each equation.

Usually two linear equations give just one common pair. The same values must work at the same time.

Simultaneous solution

A simultaneous solution satisfies every equation in a system. Write a two-variable solution as an ordered pair $(x,y)$ . An equation is linear when unknowns occur only to the first power and are not multiplied together: $3x+2y=16$ is linear; $xy=16$ is not.

Elimination makes coefficients match so that adding or subtracting removes one unknown. Keep another equation as well: a single eliminated equation does not determine both coordinates.

Worked example 1: opposite signs

Question. Solve $3x+2y=16$ and $2x-3y=-11$ .

Plan. Make the coefficients of $y$ opposite, add and back-substitute.

\begin{aligned} 9x+6y&=48 &&\text{first equation times 3}\\ 4x-6y&=-22 &&\text{second equation times 2}\\ 13x&=26 &&\text{add}\\ x&=2,\qquad y=5 &&\text{solve and use }3x+2y=16 \end{aligned}

Why this step? Opposite $y$ -terms cancel, leaving one unknown with a non-zero coefficient.

Answer. $(x,y)=(2,5)$ .

Check. $6+10=16$ and $4-15=-11$ .

Now make the matched coefficients the same sign. The cancellation requires subtraction.

Worked example 2: the same signs

Question. Solve $3x+2y=16$ and $2x+3y=19$ .

Plan. Match the coefficients of $y$ , subtract the equations and recover $y$ .

\begin{aligned} 9x+6y&=48 &&\text{first equation times 3}\\ 4x+6y&=38 &&\text{second equation times 2}\\ 5x&=10 &&\text{subtract the second from the first}\\ x&=2,\qquad y=5 &&\text{solve and back-substitute} \end{aligned}

Why this step? Adding would leave $12y$ ; subtraction is what removes equal terms.

Answer. $(x,y)=(2,5)$ .

Check. $3(2)+2(5)=16$ and $2(2)+3(5)=19$ .

The opening systems show what happens if elimination cancels both unknowns. A contradiction means no solution; an identity means the equations have repeated the same condition. Geometrically, those are distinct parallel lines and the very same line, respectively. Neither elimination nor these classifications are supplied in the formula sheet.

Watch out

Wrong: $3x+2y=16$ becomes $9x+6y=16$ after multiplying by $3$ . Multiply the right side too: it must be $48$ .

Use coefficients that cancel immediately in this check.

Substitution and fractional equations

Substitution uses an expression from one equation in place of an equal variable in the other. A coefficient of $1$ often makes this efficient.

Substitution into a linear equation

If $y=\mathrm g(x)$ , the simultaneous system with $ax+by=c$ reduces to

ax+b\mathrm g(x)=c,

then $y=\mathrm g(x)$ . Retain the domains of both equations.

Not in the formula sheet: learn it.

Write brackets around the inserted expression. They show which terms a coefficient or minus sign acts on.

Worked example 3: substituting after a minus sign

Question. Solve $y=3x-2$ and $2x-y=1$ .

Plan. Replace $y$ by its complete expression and solve for $x$ .

\begin{aligned} 2x-(3x-2)&=1 &&\text{substitute for }y\\ -x+2&=1 &&\text{expand the negative bracket}\\ x&=1,\qquad y=1 &&\text{solve and back-substitute} \end{aligned}

Why this step? The minus sign belongs to all of $3x-2$ , so its constant term becomes positive.

Answer. $(x,y)=(1,1)$ .

Check. $1=3(1)-2$ and $2(1)-1=1$ .

Now change the presentation to fractions and brackets. Clearing them exposes a simple substitution again.

Worked example 4: a fractional equation

Question. Solve $\dfrac{x+3y}{2}=5$ and $3(x-y)=6$ .

Plan. Simplify both equations, make $x$ the subject and substitute.

\begin{aligned} x+3y&=10,\qquad x-y=2 &&\text{clear the constants}\\ x&=y+2 &&\text{rearrange the second equation}\\ (y+2)+3y&=10 &&\text{substitute in the first}\\ y&=2,\qquad x=4 &&\text{solve and recover }x \end{aligned}

Why this step? Constant denominators can be cleared everywhere, without introducing excluded values.

Answer. $(x,y)=(4,2)$ .

Check. $(4+6)/2=5$ and $3(4-2)=6$ .

For linear equations, either elimination or substitution is valid. Choose the route that keeps signs and fractions manageable. If a variable occurs in a denominator, preserve its non-zero restriction after clearing it.

Watch out

Wrong: $2x-(3x-2)=2x-3x-2$ . Distribute the minus sign to both terms: $2x-3x+2$ .

Use a fractional equation with a new coefficient in the numerator.

Linear and quadratic systems

With one linear and one quadratic equation, reduce to one unknown just as before. This time the reduced equation may have two roots, so carry both branches through.

You already know

For a quadratic equal to zero, factorisation or the quadratic formula finds the real roots. Each root must then be put into the linear equation to recover its partner.

A two-variable quadratic can contain squares or a mixed $xy$ term. Replacing one variable by a linear expression makes these manageable.

Worked example 5: retaining both pairs

Question. Solve $y=x-1$ and $x^2+y^2=13$ .

Plan. Substitute for $y$ , solve the resulting quadratic and pair each root correctly.

\begin{aligned} x^2+(x-1)^2&=13 &&\text{substitute}\\ 2x^2-2x-12&=0 &&\text{expand and collect}\\ (x-3)(x+2)&=0 &&\text{divide by 2 and factorise}\\ x=3&\Rightarrow y=2 &&\text{first branch}\\ x=-2&\Rightarrow y=-3 &&\text{second branch} \end{aligned}

Why this step? A root belongs with the height obtained from that same root, not the other branch's height.

Answer. $(3,2)$ and $(-2,-3)$ .

Check. $9+4=13$ and $4+9=13$ ; both pairs also satisfy $y=x-1$ .

The next quadratic has a mixed term instead of a sum of squares. Only its form changes the substitution decision.

Worked example 6: choosing the shorter substitution

Question. Solve $x+y=6$ and $2xy+y^2=27$ .

Plan. Use $x=6-y$ , leaving the existing square in $y$ .

\begin{aligned} 2(6-y)y+y^2&=27 &&\text{substitute for }x\\ y^2-12y+27&=0 &&\text{expand and rearrange}\\ (y-3)(y-9)&=0 &&\text{factorise}\\ y=3&\Rightarrow x=3 &&\text{first pair}\\ y=9&\Rightarrow x=-3 &&\text{second pair} \end{aligned}

Why this step? Solving for $x$ first avoids introducing an extra square of a bracket into the mixed-term expression.

Answer. $(3,3)$ and $(-3,9)$ .

Check. $2(3)(3)+9=27$ and $2(-3)(9)+81=27$ ; both pairs sum to $6$ .

Do not reject a negative coordinate just because it looks less familiar. A context may forbid negative lengths, but this unrestricted system does not. These methods and domain checks are not supplied in the formula sheet.

Watch out

Wrong: combine $x=3$ with $y=-3$ from different branches in Worked example 5. The pair $(3,-3)$ fails $y=x-1$ . Write the two ordered pairs explicitly.

Try another line through the origin with a different sum of squares.

Intersections and solution counts

The graph is a useful check on a simultaneous system: common solutions are exactly common points. A repeated root supplies one distinct point, not two different intersections.

For $y=x^2+1$ and $y=2x+c$ , equating outputs gives $(x-1)^2=c$ . The diagram compares $c=1,0,-1$ : two points, touching at one point, and no points.

Three lines y=2x+c compared with y=x^2+1, showing two intersections, tangency and no intersection

At $c=1$ , the points are $(0,1),(2,5)$ . At $c=-1$ , the square would have to be negative, so no real common point exists. Algebra establishes those conclusions beyond the visible plotting window.

Worked example 7: two intersections farther apart

Question. Find the intersections of $y=x^2+1$ and $y=2x+4$ .

Plan. Equate the outputs, complete the square and recover both heights.

\begin{aligned} x^2+1&=2x+4 &&\text{same output at a common point}\\ (x-1)^2&=4 &&\text{complete the square}\\ x&=-1\text{ or }3 &&\text{take both signs}\\ x=-1&\Rightarrow y=2 &&\text{first height}\\ x=3&\Rightarrow y=10 &&\text{second height} \end{aligned}

Why this step? A positive right side gives two different offsets from $x=1$ .

Answer. $(-1,2)$ and $(3,10)$ .

Check. The parabola gives $2,10$ , as does $2x+4$ , at those respective coordinates.

Change only the line's constant to zero. Both offsets collapse to zero.

Worked example 8: a repeated root

Question. Determine the intersection of $y=x^2+1$ and $y=2x$ .

Plan. Equate outputs and identify the repeated root.

\begin{aligned} x^2-2x+1&=0 &&\text{equate and rearrange}\\ (x-1)^2&=0 &&\text{complete the square}\\ x&=1,\qquad y=2 &&\text{one distinct pair} \end{aligned}

Why this step? The square is zero at just one value. The line touches the curve at that common point.

Answer. One intersection, $(1,2)$ .

Check. Both outputs are $2$ at $x=1$ ; the quadratic difference is positive elsewhere.

Thus the family has two intersections for $c>0$ , one for $c=0$ and none for $c<0$ . This classification is not a supplied formula: it follows from the non-negativity of a real square. Keep any stated domain restriction when counting roots.

Exam tip

“Determine” needs justification. Show the reduced quadratic and its repeated-root or discriminant condition; an apparently touching calculator graph alone is insufficient.

Use the same count argument after changing the gradient and curve height.

Quick Recap

Eliminate by adding opposite coefficients or subtracting matching ones.
Substitute a complete expression, using brackets to control signs.
A linear/quadratic system may have two real branches; recover both ordered pairs.
A repeated root gives one common point, and a negative required square gives none.

Changing the opening system's $7$ to $6$ really does turn no solutions into infinitely many: two distinct parallel lines become the same line. A line and a quadratic can instead have zero, one or two intersections.

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Ma4, Ma5 - Simultaneous equations and intersections

Eliminating a variable

Substitution and fractional equations

Linear and quadratic systems

Intersections and solution counts

Quick Recap

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