Question

If $10y=7x-4$ and $12x+18y=1$ ; find the value of $4x+6y$ and $8y-x$ .
(a) $4x+6y=\dfrac{2}{5}\text{ and }8y-x=-\dfrac{7}{6}$
(b) $4x+6y=\dfrac{6}{5}\text{ and }8y-x=-\dfrac{10}{3}$
(c) $4x+6y=\dfrac{7}{3}\text{ and }8y-x=-\dfrac{4}{7}$
(d) $4x+6y=\dfrac{1}{3}\text{ and }8y-x=-\dfrac{5}{3}$

Answer 1

Hint: We will first rearrange the terms in the both equations and we can write it as $7x-10y-4=0$ and $12x+18y-1=0$. Then we will make x term as subject variable from first equation and will substitute that value in second equation. Thus, on solving we will get the value of y. Similarly, we will find x term by substituting the value of y in the first equation. Then on getting values of x and y, we will put that in $4x+6y$ and $8y-x$ to get an answer.

Complete step-by-step answer:
Here, we are given two equations $10y=7x-4$ and $12x+18y=1$ . On taking all terms on one side, we get as
$7x-10y-4=0$ ……………….(1)
$12x+18y-1=0$ …………………(2)
Now, we will make x as variable from equation (1) and we get as
$7x=10y+4$
On dividing both side by 7, we get x as
$x=\dfrac{10y+4}{7}$ …………………(3)
Now we will put this value of x in equation (2) so, we get as
$12\left( \dfrac{10y+4}{7} \right)+18y-1=0$
On opening the brackets, we get as
$\dfrac{120y+48}{7}+18y-1=0$
On taking LCM of 7, we can write it as
$120y+48+126y-7=0$
On further solving, we get as
$246y+41=0$
On taking constant term on right side and dividing both side by 246 we get as
$y=-\dfrac{41}{246}$ …………………(4)
Now, we will substitute this value of y in equation (1). So, we get as
$7x-10\left( \dfrac{-41}{246} \right)-4=0$
On simplification, we get as
$7x+\dfrac{410}{246}-4=0$
After taking LCM, we get as
$7x+\dfrac{410-984}{246}=0$
$7x-\dfrac{574}{246}=0$
On taking constant term on right side and dividing both side by 7 we get as
$x=\dfrac{574}{7\times 246}=\dfrac{82}{246}$
On further dividing, we get x as
$x=\dfrac{41}{123}$ ……………(5)
Now, we have values of x and y so we will put that value in $4x+6y$ . We get as
$4x+6y=4\left( \dfrac{41}{123} \right)+6\left( -\dfrac{41}{246} \right)$
On further simplification, we get as
$4x+6y=\left( \dfrac{164}{123} \right)-\left( \dfrac{246}{246} \right)=\dfrac{164}{123}-1$
On taking LCM and solving we get as
$4x+6y=\dfrac{164-123}{123}=\dfrac{41}{123}$
$4x+6y=\dfrac{1}{3}$ ……………………(6)
Similarly, we will put value of x and y in equation $8y-x$ so we get as
$8y-x=8\left( -\dfrac{41}{246} \right)-\dfrac{41}{123}$
On simplification, we get as
$8y-x=8\left( -\dfrac{1}{6} \right)-\dfrac{1}{3}$
$8y-x=-\dfrac{8}{6}-\dfrac{1}{3}=\dfrac{-4-1}{3}=-\dfrac{5}{3}$ …………………..(7)
Hence, option (d) is the correct answer.

Note: Here, we have using substitution method in finding value of x, y instead we can use elimination method given as we have to make any one term either x or y in both equations same and then perform operation on them. Here, we are given that $7x-10y-4=0$ and $12x+18y-1=0$ , so on making x term same in equation we will multiply first equation with 12 and second with 7. So, we get as
\[\begin{align}
  & 84x-120y-48=0 \\
 & \overset{-}{\mathop{84}}\,x\overset{-}{\mathop{+}}\,126y\overset{+}{\mathop{-}}\,7=0 \\
 & \overline{0x-246y-41=0} \\
\end{align}\]
So, in solving this way we will get the value of y. Similarly, we can do for finding x terms and thus, we will get the same answer.