Make \(x\) the subject of \(y = \frac{3x + 5}{2}\). (2) Make \(r\) the subject of \(A = \pi r^2\). (2) Given that \(v = u + at\), find the value of \(a\) wh...
Question 1 Report
- Make \(x\) the subject of \(y = \frac{3x + 5}{2}\). (2)
- Make \(r\) the subject of \(A = \pi r^2\). (2)
- Given that \(v = u + at\), find the value of \(a\) when \(v = 25\), \(u = 7\) and \(t = 3\). (2)
Answer Details
This question tests rearranging formulae to change the subject, including one involving a fraction and one involving a square, and substituting into a formula to find an unknown letter.
- Multiplying both sides of \(y = \dfrac{3x+5}{2}\) by 2 gives \(2y = 3x + 5\); subtracting 5 and dividing by 3 gives \(x = \dfrac{2y-5}{3}\). [2]
- Dividing both sides of \(A = \pi r^2\) by \(\pi\) gives \(r^2 = \dfrac{A}{\pi}\); taking the square root of both sides gives \(r = \sqrt{\dfrac{A}{\pi}}\). [2]
- Substituting \(v = 25\), \(u = 7\), \(t = 3\) into \(v = u + at\): \(25 = 7 + 3a\), so \(3a = 18\), giving \(a = 6\). [2]
In each part, the operations applied to the original letter (multiply, add, square) are undone in reverse order to isolate the new subject; a square root must always come last, once every other term has already been cleared, which is why part (b) squares before it roots.
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