# Solve the following equations:$(a).\ 2y+\frac{5}{2}=\frac{37}{2}$$(b).\ 5t+28=10$$(c).\ \frac{a}{5}+3=2$$(d).\ \frac{q}{4}+7=5$$(e).\ \frac{5}{2}x=-5$$(f).\ \frac{5}{2}x=\frac{25}{4}$$(g).\ 7m+\frac{19}{2}=13$ $(h).\ 6z+10=-2$ $(i).\ \frac{3l}{2}=\frac{2}{3}$$(j).\ \frac{2b}{3}-5=3$

#### Complete Python Prime Pack

9 Courses     2 eBooks

#### Artificial Intelligence & Machine Learning Prime Pack

6 Courses     1 eBooks

#### Java Prime Pack

9 Courses     2 eBooks

To do:

We have to solve the given equations.

Solution:

(a) $2y+\frac{5}{2}=\frac{37}{2}$

Transposing $\frac{5}{2}$ to R.H.S we get,

$2y=\frac{37}{2}-\frac{5}{2}$

$2y=\frac{37-5}{2}$

$2y=\frac{32}{2}$

$y=\frac{16}{2}$

$y=8$

(b) $5t+28 =10$

Transposing 28 to R.H.S we get,

$5t=10-28$

$5t=-18$

$t=\frac{-18}{5}$

(c) $\frac{a}{5}+3=2$

Transposing 3 to R.H.S we get,

$\frac{a}{5}=2-3$

$\frac{a}{5}=-1$

Multiplying both side by 5, we get,

$\frac{a}{5}\times5=-1\times5$

$a=-5$

(d) $\frac{q}{4}+7=5$

Transposing 7 to R.H.S we get,

$\frac{q}{4}=5-7$

$\frac{q}{4}=-2$

Multiplying both side by 5, we get,

$\frac{q}{4}\times4=-2\times4$

$q=-8$

(e) $\frac{5}{2}x=-5$

Multiplying both side by 2, we get,

$\frac{5}{2}x \times2=-5\times2$

$5x=-10$

Dividing both sides by 5, we get,

$\frac{5x}{5}=\frac{-10}{5}$

$x=-2$

(f) $\frac{5}{2}x=\frac{25}{4}$

Multiplying both side by 2, we get,

$\frac{5}{2}x \times2=\frac{25}{4}\times2$

$5x=\frac{25}{2}$

Dividing both sides by 5, we get,

$\frac{5x}{5}=\frac{25}{2\times5}$

$x=\frac{5}{2}$

(g) $7m+\frac{19}{2}=13$

Transposing $\frac{19}{2}$ to the R.H.S.

$7m=13-\frac{19}{2}$

$7m=\frac{2(13)-19}{2}$

$7m=\frac{26-19}{2}$

$7m=\frac{7}{2}$

Dividing both sides by 7, we get,

$\frac{7m}{7}=\frac{7}{2\times7}$

$m=\frac{1}{2}$

(h) $6z+10=-2$

Transposing 10 to the R.H.S.

$6z=-2-10$

$6z=-12$

Dividing both sides by 6, we get,

$\frac{6z}{6}=\frac{-12}{6}$

$z=-2$

(i) $\frac{3}{2}l=\frac{2}{3}$

Multiplying both side by $\frac{2}{3}$, we get,

$\frac{3}{2}l \times \frac{2}{3}=\frac{2}{3}\times \frac{2}{3}$

$l=\frac{4}{9}$

(j) $\frac{2b}{3}-5=3$

Transposing $-5$ to R.H.S we get,

$\frac{2b}{3}=3+5$

$\frac{2b}{3}=8$

Multiplying both side by $\frac{3}{2}$, we get,

$\frac{2b}{3} \times \frac{3}{2}=8\times \frac{3}{2}$

$b=4\times3$

$b=12$

Updated on 10-Oct-2022 13:33:38