Question

Solve the following system of equations by the elimination method:

$$\begin{gathered} \left(xii\right)1.2a+0.5b=0.8 \\ 0.8a+0.07b=0.428 \end{gathered}$$

Answer 1

Solution :-

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

$$\begin{gathered} \frac{12}{10}a+\frac{5}{10}b=\frac{8}{10}------\left(1\right) \\ \frac{8}{10}a+\frac{7}{100}b=\frac{428}{1000}----\left(2\right) \end{gathered}$$

Step 2: Multiply 10 with both sides of equation (1) & Multiply 1000 with both sides of equation (2).

It becomes $$\begin{gathered} 12a+5b=8------\left(3\right) \\ 800a+70b=428----\left(4\right) \end{gathered}$$

Step 3: Multiply the equation (3) by suitable number $14$.

(coefficient of the variable ‘b’ becomes equal).

$\left(3\right)\times 14\Rightarrow 168a+70b=112----\left(5\right)$

Step 4: Subtracting the equation (5) from equation (4) to eliminate the variable ‘b’.

(The resulting equation is a linear equation in one variable).

$$\begin{gathered} 800a+70b=428-----\left(4\right) \\ 168a+70b=112-----\left(5\right) \end{gathered}$$

$\left(4\right)+\left(5\right)\Rightarrow$$$\begin{gathered} 632a+0b=316 \\ a=\frac{316}{632}=0.5 \end{gathered}$$

Step 5: Substitute the value of the variable $a=0.5$ in the given equation(3).

i.e. $12a\times 5b=8$, to get the value of the other variable ‘b’

$$\begin{gathered} 12\times 0.5+5b=8 \\ 6+5b=8 \\ 5b=2 \\ b=\frac{2}{5}=0.4 \end{gathered}$$

Hence, the value of $a=0.5$ and $b=0.4$