Question

Find the solution for the following pair of equations
$2x+8y=3$
$2x+y=\frac{5}{4}$

• A. $(1,-\frac{3}{4})$
• B. $(1,\frac{1}{4})$
• C. $(\frac{1}{4},\frac{5}{16})$
• D. $(\frac{1}{2},\frac{1}{4})$

Answer 1

The correct option is D $(\frac{1}{2},\frac{1}{4})$

Given,
$2x+8y=3\dots \left(i\right)$
$2x+y=\frac{5}{4}\dots \left(ii\right)$

Step 1: Find the value of y
$\left(i\right)-\left(ii\right)\Rightarrow 2x+8y-(2x+y)=3-\frac{5}{4}\Rightarrow 8y-y=3\times \frac{4}{4}-\frac{5}{4}\Rightarrow 7y=\frac{12}{4}-\frac{5}{4}\Rightarrow 7y=\frac{7}{4}$
Dividing both the sides by $7,$ we get
$\Rightarrow y=\frac{1}{4}$

Step 2: Find the value of x
Putting $y=\frac{1}{4}$ in $\left(i\right),$ we get
Equation $\left(i\right)\to 2x+8\left(y\right)=3$
$\Rightarrow 2x+8\times \frac{1}{4}=3\Rightarrow 2x+2=3$
Subtracting both the sides by 2, we get
$\Rightarrow 2x=3-2\Rightarrow 2x=1$
Dividing both the sides by $2,$ we get
$\Rightarrow x=\frac{1}{2}$
$\therefore x=\frac{1}{2},y=\frac{1}{4}$
$\therefore$ Required solution for the given equations is $(\frac{1}{2},\frac{1}{4}).$