Edexcel IGCSE Math B 3.I linear and quadratic simultaneous equations

Practise simultaneous-equation questions by substituting the linear equation, forming a quadratic and finding both solution pairs.

Exam points

Substitute the linear equation into the quadratic relationship to form one equation.
Solve the resulting quadratic and pair each x-value with the matching y-value.
Show clear algebraic working for simultaneous systems with squared terms or products.

[Maximum number: 5]

Figure 2 shows two shapes A and B.
All the lengths are in metres.

Figure 2

The corners of shape A are all right angles.
The area of shape A is $400 \mathrm{~m}^{2}$
Shape B is triangle PQR.
The perimeter of shape A is equal to the perimeter of shape B.

Find the value of x and the value of y.
Show clear algebraic working.

From part (a):
$4y^{2}-x^{2}=200,\qquad 2y=2x+5$
so
$y=x+\frac{5}{2}.$

Substitute into the area equation:
$4\left(x+\frac{5}{2}\right)^{2}-x^{2}=200$ $3x^{2}+20x-175=0$ (3x+35)(x-5)=0
Since the length is positive,
x=5.
Then
$y=x+\frac{5}{2}=7.5.$

Final answer:
$x=5,\qquad y=7.5$

Marks: M1 for substituting one equation into the other. M1 for rearranging correctly to a three-term quadratic. M1 dependent on previous method mark for solving their quadratic equation using a correct method. A1 for one correct result for x or y, ignoring a negative value for this mark. A1 dependent on method marks for x=5 and y=7.5, positive values only.

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3.I Linear and quadratic simultaneous equations question 1