Free practice · 5 questions
Practise rearranging equations, solving quadratics and turning a word problem into algebra.
These are original InfinitySheets questions. Try each one first, then open the worked solution to check your method.
Solve 3(2x − 5) = 4x + 7.
Expand the left side to get 6x − 15 = 4x + 7. Subtract 4x from both sides, then add 15 to both sides. This gives 2x = 22, so x = 11. Check by substituting 11. Both sides equal 51.
Two notebooks and three pens cost ₹96. Three notebooks and two pens cost ₹114. Find the price of one notebook and one pen.
Let n be the notebook price and p the pen price. Then 2n + 3p = 96 and 3n + 2p = 114. Multiply the first equation by 2 and the second by 3 to get 4n + 6p = 192 and 9n + 6p = 342. Subtract to get 5n = 150, so n = 30. Substitute into the first equation to get 3p = 36, so p = 12. A notebook costs ₹30 and a pen costs ₹12.
Solve 2x² − 7x + 3 = 0.
Find two numbers that multiply to 6 and add to −7. They are −6 and −1. Rewrite as 2x² − 6x − x + 3 = 0, then factorise to (2x − 1)(x − 3) = 0. Either factor can be zero, so x = ½ or x = 3.
Make r the subject of V = πr²h, where r is a radius and h > 0.
Divide both sides by πh to isolate r². Then take the square root. Because a radius is nonnegative, r = √(V ÷ (πh)). The negative root would solve the squared equation but cannot represent a radius.
A rectangle has perimeter 34 cm and area 60 cm². Find its side lengths.
If one side is x cm, the other is 17 − x cm, since the two lengths add to half the perimeter. Then x(17 − x) = 60, which rearranges to x² − 17x + 60 = 0. Factorise as (x − 5)(x − 12) = 0. The roots describe the same rectangle with its sides swapped, so the side lengths are 5 cm and 12 cm.
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