Solve the following equations:
$(a).\ 2y+\frac{5}{2}=\frac{37}{2}$
$(b).\ 5t+28=10$
$(c).\ \frac{a}{5}+3=2$
$(d).\ \frac{q}{4}+7=5$
$(e).\ \frac{5}{2}x=-5$
$(f).\ \frac{5}{2}x=\frac{25}{4}$
$(g).\ 7m+\frac{19}{2}=13$
$(h).\ 6z+10=-2$
$(i).\ \frac{3l}{2}=\frac{2}{3}$
$(j).\ \frac{2b}{3}-5=3$
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,
$\frac{2b}{3} \times \frac{3}{2}=8\times \frac{3}{2}$
$b=4\times3$
$b=12$
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