make a the subject of p = (3a+5)/(4-a)

so step 1) remove the denominator by multiplying (4-a) by p. Therefore obtaining p(4-a) = 3a+5. step 2) multiply out to get 4p-ap = 3a+5 step 3) have all the a's on one side, and non a's on other side. therefore: 4p-5 = 3a+ap step 4) factor out the a's. 4p-5 = a(3+p) step 5) make a = (4p-5)/(3+p) 

JW

Related Maths GCSE answers

All answers ▸

A point lies on a circles diameter such that the distance from the point to the edge of circle is 4 times the distance from the point to the centre. What is the circles area in cm^2 if the distance from the point to edge is 5cm?


Solve the following simultaneous equations: (1) 2y + x = 8, (2) 1 + y = 2x


The equation 5x^2 + px + q = 0, where p and q are constants, has roots t and t+4. Show that p^2 = 20q + 400.


Work out 3 and 1/2 divided by 2 and 4/5