Question

Express the following as algebraic equations and solve
(a) Twice a number increased by 7 is 13. What is the number?
(b) Seven times a number decreased by 4 is 10. Find the number.

Answer 1

(a) Let the number be x
$\therefore$ Twice the number =2x
$\therefore$ 2x+7=13

x	L.H.S = 2x + 7	R.H.S = 13
1	$2\times 1+7=9$	13
2	$2\times 2+7=11$	13
3	$2\times 3+7=13$	13

Thus, for x=3, L.H.S. = R.H.S. So the required number is 3

(b) Let the numbers be x
$\therefore$ 7x - 4 = 10

x	L.H.S = 7x - 4	R.H.S = 10
1	$7\times 1-4=3$	10
2	$7\times 2-4=10$	10

Here for x=2, L.H.S. = R.H.S.
$\therefore$ The required number is 2