Question

Solve the following pairs of linear (simultaneous) equation by the method of elimination:$0.2x+0.1y=25$, $2(x-2)-1.6y=116$

• A. $x=100$, $y=50$
• B. $x=80$, $y=30$
• C. $x=-78$, $y=35$
• D. $x=-65$, $y=-75$

Answer 1

The correct option is A $x=100$, $y=50$

Multiply the equation $0.2x+0.1y=25$ by $10$ and the equation $2(x-2)-1.6y=116$ can be rewritten as $2x-1.6y=120$. Then we get the following equations:

$2x+y=\mathrm{250.........}\left(1\right)$

$2x-1.6y=\mathrm{120.........}\left(2\right)$

Subtract Equation (2) from Equation (1) to eliminate $x$, because the coefficients of $x$ are the same. So, we get

$(2x-2x)+(y+1.6y)=250-120$

i.e. $2.6y=130$

i.e. $y=50$

Substituting this value of $y$ in the equation $2x-1.6y=120$, we get

$2x-80=120$

i.e. $2x=200$

i.e. $x=100$

Hence, the solution of the equations is $x=100,y=50$.