Given a^2 < 4 and a+2b = 8. Work out the range of possible values of b. Give your answer as an inequality.

For this question, we want to start by finding the possible values of a from the first equation and then using that to give us information about b from the second equation.

Now, the condition that a^2 < 4 actually gives us two pieces of information (a way to remember this is that for inequalities, squared gives us two things). We have that a < 2 but also that a >  -2 since for example, (-3)^2 = 9 which is larger than 4. Therefore, we have -2 < a < 2.

Rearranging the second equation for b, we get b = (8-a)/2. Therefore, putting in our values for a, we get that the maximum value of b is (8-(-2))/2 = 10/2 = 5 and the minimum value of b is (8-2)/2 = 6/2 = 3. Since the inequality for a does not include 2 and -2, we do not include 3 and 5 for b and so we get the answer 3 < b < 5.

FS

Related Further Mathematics GCSE answers

All answers ▸

What is the distance between two points with x-coordinates 4 and 8 on the straight line with the equation y=(3/4)x-2


Find the coordinates of the minimum/maximum of the curve: Y = 8X - 2X^2 - 9, and determine whether it is a maximum or a minimum.


How can a system of two linear equations be solved?


A straight line passes trough the points A(-4;7); B(6;-5); C(8;t). Use an algebraic method to work out the value of t.