Answer:
[tex](a)\ y = 4\ or\ y = -\frac{1}{2}[/tex]
[tex](b)\ x = 2 \ or\ x = -\frac{1}{4}[/tex]
Step-by-step explanation:
Solving (a)
[tex]2y^2 - 7y - 4 = 0[/tex]
Expand
[tex]2y^2 +y - 8y - 4 = 0[/tex]
Expand
[tex]y(2y + 1) -4(2y + 1) = 0[/tex]
Factor out 2y + 1
[tex](y -4)(2y + 1) = 0[/tex]
Split
[tex]y - 4 =0\ or\ 2y + 1 = 0[/tex]
Solve:
[tex]y = 4\ or\ y = -\frac{1}{2}[/tex]
Solving (b):
[tex]2(2x)^2 - 7(2x) - 4 = 0[/tex]
Compare the above with the equation in (a)
[tex]2y^2 - 7y - 4 = 0[/tex]
We have:
[tex]2x = y[/tex]
So:
[tex]2x = 4 \ or\ 2x = -\frac{1}{2}[/tex]
Solve for x in both cases
[tex]x = 2 \ or\ x = -\frac{1}{4}[/tex]
