Simultaneous equationsEdexcel A-Level Maths: Revision notes

Section 1

Simultaneous equations by elimination

Simultaneous equations must both be true at once. For two linear equations you can use elimination: multiply to make one pair of coefficients equal, then add or subtract. Example: $2x-3y=6$ and $x+2y=-1$ . Multiply the second by 2: $2x+4y=-2$ . Subtract: $-7y=8$ , so $y=-\frac87$ . Substitute back for $x$ . Always check by substituting both values into the original equations. A solution is a pair of values.

Key termssimultaneous equationselimination

Common mistake

Adding when the signs of the coefficients are the same (or subtracting when they differ). Check the sign of the term you want to cancel.

Section 2

Substitution: one linear and one quadratic

When one equation is quadratic, use substitution: rearrange the linear equation to make a variable the subject, and substitute into the quadratic. Example: $y=2x+3$ and $y=x^2-4x+8$ . Then $x^2-4x+8=2x+3$ , so $x^2-6x+5=0$ , $(x-1)(x-5)=0$ . So $x=1,y=5$ or $x=5,y=13$ . A linear and a quadratic equation usually give two pairs of solutions, which represent the two points where a line meets a curve.

Key termssubstitutionpair of solutions

Common mistake

Finding both $x$ values but pairing them with the wrong $y$ . Substitute each $x$ into the linear equation to find its $y$ .

Section 3

Harder pairs

When the quadratic has an $x^2$ and a $y^2$ term, make the variable with fewer fractions the subject. Example: $2x-3y=6$ and $x^2-y^2+3x=50$ . From the first, $x=3+\frac32y$ . Substituting gives $5y^2+54y-128=0$ , so $(5y+64)(y-2)=0$ . Then $y=2$ with $x=6$ , or $y=-\frac{64}{5}$ with $x=-\frac{81}{5}$ . Alternatively eliminate $y$ to get $5x^2+51x-486=0$ . Tidy by multiplying through to remove fractions. Always find the other variable from the linear equation.

Key termsquadratic in one variable

Exam tip

Clear fractions early by multiplying every term by the common denominator.

Section 4

Powers of 2 in the unknowns

Some pairs have the unknowns as indices. Write everything in the same base and use the index laws to turn them into linear equations. Example: $2^x\times4^y=32$ and $\frac{8^x}{2^y}=2$ . Then $4^y=2^{2y}$ and $8^x=2^{3x}$ , so $x+2y=5$ and $3x-y=1$ . Solving: $x=1$ , $y=2$ . In other questions only one equation has the index. Use substitution as usual, remembering $2^{a+b}=2^a2^b$ .

Key termsindex lawsame base

Common mistake

Adding the bases: $2^x\times4^y\neq8^{x+y}$ . Convert 4 to $2^2$ first.

Section 5

Line meets curve: the discriminant link

Substituting a line into a curve gives a quadratic in one variable whose discriminant tells you how many intersections there are.

$b^2-4ac>0$ : line crosses the curve at two points.
$b^2-4ac=0$ : line is a tangent (touches at one point).
$b^2-4ac<0$ : no intersection. Example: $y=2x+k$ and $y=x^2-4x+8$ give $x^2-6x+8-k=0$ . A tangent needs $36-4(8-k)=0$ , so $k=-1$ , $x=3$ , $y=5$ . Another example: a rectangle with $x+y=17$ and $x^2+y^2=169$ gives $x^2-17x+60=0$ , so the sides are 5 and 12.

Key termstangentdiscriminant

Exam tip

If two equations give a repeated root, there is exactly one solution: the line is a tangent.

That's the notes covered.

Carry on to the next subtopic.

Exam questions on Simultaneous equations

The line $y=x+1$ meets the curve $y=x^2-3x-4$ at two points.
The line $y=x+c$ is a tangent to the curve $y=x^2-3x-4$ . Find the value of $c$ .2 marks
A rectangle has perimeter 34 cm and its diagonal has length 13 cm. Its sides have lengths $x$ cm and $y$ cm.
Without solving a quadratic equation, find the area of the rectangle.2 marks
Real numbers $x$ and $y$ satisfy $2^{x}\times4^{y}=32$ and $\frac{8^{x}}{2^{y}}=2$ .
Show that $x+2y=5$ and $3x-y=1$ .3 marks

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Written by the Exaim team, led by Shaun Daswani (Head of Upper Secondary, Improve ME Institute; MSc Financial Mathematics, Imperial College London; BSc, UCL) and Jason Daswani (operational lead, Improve ME Institute; LSE).