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Maths
GCSE

Here is a right-angled triangle (base = 8cm and height = 9cm) and a rectangle (length = 16cm). The area of the rectangle is 6 times the area of the triangle. Work out the width of the rectangle.

Area of triangle is 0.5 x base x height.
0.5 x 8 x 9 = 36
Area of rectangle is 6 x 36 = 216
Area of rectangle is also 16 x width
16 x width = 216
(Divide both sides by 16)
w...

Answered by Harry W. Maths tutor
11146 Views

Solve the simultaneous equations 3x+2y=4 and 4x+5y=17 for x and y

Rearrange one of the equations to get x equals an equation in terms of y.3x+2y=4Take 2y away from both sides3x=4-2yDivide both sides by 3x=(4-2y)/3Substitute this into the other equation in place of x4((4...

Answered by Jack C. Maths tutor
6414 Views

Solve the equation (6/x-2)-(6/x+1) =1

1 - Multiply each part of equation by denominators to multiply out the fractions. {(6(x-2))/x-2} - {(6(x-2))/x+1} = 1(x-2). This would simply to 6 - {(6(x-2))/x+1} = x - 2. We then do the same for the oth...

Answered by Aman K. Maths tutor
7978 Views

A is the point with coordinates (2,7) and B is the point with coordinates (8,10). Find the coordinates of the midpoint of AB.

1 - Draw a sketch graph of where the points would be to help visualise the midpoint. 2 - Work out the midpoint for each axis. For example, for x, it would be (0.5*(2+8)) = 5 and for y, it would be (0.5*(7...

Answered by Aman K. Maths tutor
4661 Views

(Q20 Non-Calculator paper, Higher Tier) Solve algebraically the simultaneous equations: x^2 + y^2 = 25 and y – 3x = 13

We have a quadratic equation (unknowns are raised to the power of 2) and a liner equation (power 1). We will use the linear equation to express one of the variables, e.g. y, in terms of t...

Answered by Darena S. Maths tutor
4116 Views

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