Explanation:
we need to solve both equations for x ?
1.
(5 + 3x)/10 - 2(x - 1)/8 = 1
first, let's bring both fractions to the same denominator.
the LCM (last common multiple) of 10 and 8 is 40 (the smallest number that can be divided by 10 and by 8 without remainder).
10×4 = 40
8×5 = 40
so,
4×(5 + 3x)/40 - 5×2×(x - 1)/40 = 1
then we simply multiply both sides by 40 to eliminate the fractions altogether :
4×(5 + 3x) - 5×2×(x - 1) = 40
20 + 12x - 10x + 10 = 40
2x + 30 = 40
2x = 10
x = 5
2.
(x² + x - 1)/(x² + 5x + 4) + 1/(x+4) = (x - 1)/(x + 1)
x² + 5x + 4 = x² + 8x + 16 - 3x - 12 = (x + 4)² - 3x - 12 =
= (x + 4)² - 3(x + 4) = (x + 4)((x + 4) - 3)
so,
(x²+x-1)/((x+4)((x+4) - 3)) + 1/(x+4) = (x-1)/(x+1)
let's multiply both sides by (x+4) :
(x²+x-1)/((x+4) - 3) + 1 = (x+4)(x-1)/(x+1)
(x²+x-1)/(x+1) + 1 = (x+4)(x-1)/(x+1)
now we multiply both sides by (x+1) :
x² + x - 1 + (x + 1) = (x + 4)(x - 1)
x² + 2x = (x + 4)(x - 1) = x² + 3x - 4
2x = 3x - 4
0 = x - 4
x = 4