Simultaneous equations


Calculators are allowed for all questions unless indicated by the question. There are 5 questions for this topic. The difficulty increases with the question number.

Question 1: (Non-calculator)

Find the values of y and x for the following simultaneous equation:

$x = y +4$

$y = 4x + 2$ (3 marks)

Substitute $x = y+4$ into $y=4x+2$

$y=4(y+4) + 2$

$y=4y + 18$ (1 mark)

$3y = -18$

$y = -6 $ (1 mark)

$\therefore x = -6 +4 = -2$ (1 mark)

Question 2:

Solve the following simultaneous equation:

$y = x +6$

$y = x^2$ (5 marks)

since $y=x^2$, we can substitute $x^2$ into $y=x+6$

$x^2 = x + 6$

$x^2 – x – 6 = 0$ (1 mark)

$x = 3, x= -2 $ (1 mark)

Using $x=3$, $y = 3^2 = 9$ (1 mark)

Using $x = -2$, $y = (-2)^2 = 4$ (1 mark)

$\therefore x =3, y = 9,$ or $x = -2, y = 4$ (1 mark)

Do not award final mark if values of $x$ and $y$ are not in the correct pairs

Question 3:

Solve the following simultaneous equation:

Joshua buys 3 apples and 2 oranges for a total price of £5.95. He sells 1 of his apples and both oranges to James for a total of £3.55 where each item was sold for the same price he bought it at.

A. Calculate the price of an apple and an orange to the nearest penny. (5 marks)

B. Demonstrate this relationship on a graph, labelling the co-ordinates of the point of intersection (4 marks)

The following problem can be described with simultaneous equations, where $x$ = apple, and $y$ = orange

Equation 1: $3x + 2y = 5.95$

Equation 2: $x + 2y = 3.55$ (1 mark for correctly identifying the problem as a simultaneous equations)

Multiply the second equation by 3 so both equations have $3x$:

Equation 2: $3x +6y = 10.65$

Subtract equation 1 from equation 2 to eliminate $x$:

Left hand side: $(3x +6y) – (3x + 2y) = 4y$

Right hand side: $10.65 – 5.95 = 4.70$ (2 marks for eliminating $x$ or $y$)

$4y = 4.70$

$y = £1.175 = £1.18$ (1 mark)

$3x = 10.65 – 6(1.175) = 3.60 $ (note that you must use the exact value of y when calculating the value of x)

$x = £1.20$ (1 mark)

Award full marks for any alternative method including substitution

Rearrange each equation in the form $y=mx+c$ so you can draw it.

$3x + 2y = 5.95$

Equation 1: $y = -1.5x + 2.975$ (1 mark)

$x + 2y = 3.55$

Equation 2: $y = -x + 1.775$ (1 mark)

Sketch a graph of these 2 equations by testing $y=0$ and $x=0$ and plotting 2 points of reference for each equation and drawing a straight line between them.

(2 marks for drawing correct graph, allow 1 mark if co-ordinates are incorrect or missing)

Question 4:

A. Given the shape is a rectangle, calculate the value of $y$ and $x$ (4 marks)

B. Calculate the area of the rectangle (2 marks)

Since the opposite sides are the same length, $9-y=2x+1$ and $4y-2 = 4x$ (1 mark)

Multiply the first equation by 2 to eliminate $x$ when subtrated from equation 2:

$18-2y = 4x + 2$ (1 mark)

Subtract equation 2 from equation 1:

$20 – 6y = 2$

$y = 3$ (1 mark)

$9-3 = 2x +1$, $x = 2.5$ (1 mark)

Area = $(9-y) \times 4x$ (1 mark)

$(9-3) \times (4\times2.5) = 60$ (1 mark)

Question 5:

Given the following equations and that they intersect twice:

$y= x^2 – 4$

$y = 4x + 8$

Calculate the points of intersection (4 marks)

Intersection happens when $4x +8 = x^2 – 4$ (1 mark)

$x^2 -4x -12 = 0$

$x = -2, x=6$ (1 mark)

Substituting values back into $y = x^2 -4$:

$y = (-2)^2 -4 = 0$

$y = 6^2 -4 = 32$

$x = -2, y= 0$ and $x=6, y = 32$ (2 marks)

Allow 1 mark if pairs are incorrect