Question

Solve the following system of equations by the elimination method:

$$\begin{gathered} \left(xi\right)2x+0.4y=1.2 \\ 3.4-0.02y=0.01 \end{gathered}$$

Answer 1

Solution :-

Step 1: Convert the decimal into fractional forms of the given equations.

$$\begin{gathered} 2x+\frac{4}{10}y=\frac{12}{10}------\left(1\right) \\ \frac{34}{10}x-\frac{2}{100}y=\frac{1}{100}----\left(2\right) \end{gathered}$$

Multiply 10 with both sides of equation (1) & Multiply 100 with both sides of equation (2)

it becomes $$\begin{gathered} 20x+4y=12----\left(3\right) \\ 340x-2y=1----\left(4\right) \end{gathered}$$

Step 2: Multiply the equation (4) by suitable number( $2$). The coefficient of the variable ‘y’ becomes equal.

$\left(4\right)\times 2\Rightarrow 680x-4y=2----\left(5\right)$

Step 3: Adding equation (3) and the new equation obtained(5) to eliminate the variable ‘y’. And the resulting equation is a linear equation in one variable.

$$\begin{gathered} 20x+4y=12-----\left(3\right) \\ 680x-4y=2-----\left(5\right) \end{gathered}$$

$\left(3\right)+\left(5\right)\Rightarrow$

$$\begin{gathered} 700x=14 \\ x=\frac{14}{700} \\ x=0.02 \end{gathered}$$

Step 4: Substitute the value of the variable $x=0.02$ in the given equation(3)

i.e. $20x+4y=12$, to get the value of the other variable ‘y’

$$\begin{gathered} 20\times 0.02+4y=12 \\ 0.04+4y=12 \\ 4y=12-0.04 \\ 4y=11.6 \\ y=\frac{11.6}{4}=2.9 \end{gathered}$$

Hence, the value of $x=0.02$ and $y=2.9$