Question

Please solve the linear equations of the A, B, and D don't write of C please don't write I don't know and Don't write other subjects question or I will report to brainly.​​​

Answer 1

$\large\underline{\sf{Solution-a}}$

Given linear equation is

$\rm :\longmapsto\:0.6a - 15 = - 1.2a + 3$

On transposition we get

$\rm :\longmapsto\:0.6a + 1.2a = 15 + 3$

$\rm :\longmapsto\:1.8a = 18$

$\rm :\longmapsto\:a = \dfrac{18}{1.8} = \dfrac{18 \times 10}{18} = 10$

$\bf\implies \:a = 10$

Verification :-

Consider LHS

$\rm :\longmapsto\:0.6a - 15$

Put value of a = 10, we get

$\rm \: = \: \:0.6 \times 10 - 15$

$\rm \: = \: 6 - 15$

$\rm \: = \: - 9$

Consider, RHS

$\rm :\longmapsto\: - 1.2a + 3$

On substituting a = 10, we get

$\rm \: = \: \: - 1.2 \times 10 + 3$

$\rm \: = \: \: - 12 + 3$

$\rm \: = \: \: - 9$

Hence, Verified

$\large\underline{\sf{Solution-b}}$

Given linear equation is

$\rm :\longmapsto\:\dfrac{4a}{5} - \dfrac{1}{2} = \dfrac{2a}{3} - \dfrac{3}{4}$

On transposition we get

$\rm :\longmapsto\:\dfrac{4a}{5} -\dfrac{2a}{3} = \dfrac{1}{2} - \dfrac{3}{4}$

$\rm :\longmapsto\:\dfrac{12a - 10a}{15} = \dfrac{2 - 3}{4}$

$\rm :\longmapsto\:\dfrac{2a}{15} = - \dfrac{1}{4}$

$\bf\implies \:a = - \: \dfrac{15}{8}$

Verification :-

$\rm :\longmapsto\:\dfrac{4a}{5} - \dfrac{1}{2} = \dfrac{2a}{3} - \dfrac{3}{4}$

On substituting the value of a, we get

$\rm :\longmapsto\: - \dfrac{4}{5} \times \dfrac{15}{8} - \dfrac{1}{2} = - \dfrac{2}{3} \times \dfrac{15}{8} - \dfrac{3}{4}$

$\rm :\longmapsto\: - \dfrac{3}{2} - \dfrac{1}{2} = - \dfrac{5}{4} - \dfrac{3}{4}$

$\rm :\longmapsto\: - \dfrac{4}{2} = - \dfrac{8}{4}$

$\rm :\longmapsto\: - 2 = - 2$

Hence, Verified

$\large\underline{\sf{Solution-d}}$

Given Linear equation is

$\rm :\longmapsto\:\dfrac{c - 3}{3} + \dfrac{c + 5}{5} = \dfrac{c - 2}{2}$

$\rm :\longmapsto\:\dfrac{5(c - 3) + 3(c + 5)}{15} = \dfrac{c - 2}{2}$

$\rm :\longmapsto\:\dfrac{5c - 15 + 3c +15}{15} = \dfrac{c - 2}{2}$

$\rm :\longmapsto\:\dfrac{8c}{15} = \dfrac{c - 2}{2}$

$\rm :\longmapsto\:16c = 15c - 30$

$\rm :\longmapsto\:16c - 15c = - 30$

$\bf\implies \:c = - \: 30$

Verification

$\rm :\longmapsto\:\dfrac{c - 3}{3} + \dfrac{c + 5}{5} = \dfrac{c - 2}{2}$

On substituting the value of c = - 30, we get

$\rm :\longmapsto\:\dfrac{ - 30- 3}{3} + \dfrac{ - 30 + 5}{5} = \dfrac{ - 30 - 2}{2}$

$\rm :\longmapsto\:\dfrac{ - 33}{3} + \dfrac{ - 25}{5} = \dfrac{ - 32}{2}$

$\rm :\longmapsto\: - 11 - 5 = - 16$

$\rm :\longmapsto\: - 16 = - 16$

Hence, Verified