How do you solve $\frac{1}{5} (4 n + 1) = \frac{1}{8} (7 n - 2)$ $\frac { 1} { 5} ( 4n + 1) = \frac { 1} { 8} ( 7n - 2)$ ?

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Answer 1 · 2018-04-13T16:31:46

$n = 6$ $n=6$

Multiply both sides by 40 (5 and 8) this will cancel out the fractions and leave a linear equation to solve.

$40 \times \frac{1}{5} (4 n + 1) = 40 \times \frac{1}{8} (7 n - 2)$ $40xx1/5(4n+1)=40xx1/8(7n-2)$ .
Cancelling leaves

$8 (4 n + 1) = 5 (7 n - 2)$ $8(4n+1)=5(7n-2)$
Expanding the brackets leaves

$32 n + 8 = 35 n - 10$ $32n+8=35n-10$
Subtract $32 n$ $32n$ from both sides

$8 = 3 n - 10$ $8=3n-10$
Add 10 to both sides

$18 = 3 n$ $18=3n$

$n = 6$ $n=6$

How do you solve \frac { 1} { 5} ( 4n + 1) = \frac { 1} { 8} ( 7n - 2)15(4n+1)=18(7n−2)\frac { 1} { 5} ( 4n + 1) = \frac { 1} { 8} ( 7n - 2)?