Question

Solve the following system of equation by using the method of elimination by equating the coefficients: x/10+y/5+1=15 and x/8+y/6=15

Answer 1

Answer:

This is the answer for this question.

Answer 2

Solution :

Given :

x/10 + y/5 + 1 = 15 & x/8 + y/6 = 15

$\underline{\bf{Explanation\::}}}$

$\mapsto\tt{\dfrac{x}{10} +\dfrac{y}{5} +1=15}\\\\\mapsto\tt{\dfrac{x+2y+10}{10} =15}\\\\\mapsto\tt{x+2y+10=150}\\\\\mapsto\tt{x+2y=150-10}\\\\\mapsto\tt{x+2y=140........................(1)}$

&

$\mapsto\tt{\dfrac{x}{8} +\dfrac{y}{6} =15}\\\\\mapsto\tt{\dfrac{3x+4y}{24} =15}\\\\\mapsto\tt{3x+4y=360........................(2)}$

Now, multiply in equation (1) by 2, we get;

$\mapsto\tt{2(x+2y=140)}\\\\\mapsto\tt{2x+4y=280..................(3)}$

So, subtracting equation (2) and equation (3),we get;

$\mapsto\tt{3x-2x\cancel{+4y-4y}=360-280}\\\\\mapsto\bf{x=80}$

Putting the value of x in equation (3),we get;

$\mapsto\tt{2(80) + 4y=280}\\\\\mapsto\tt{160+4y=280}\\\\\mapsto\tt{4y=280-160}\\\\\mapsto\tt{4y=120}\\\\\mapsto\tt{y=\cancel{120/4}}\\\\\mapsto\bf{y=30}$

Thus;

The value of x and y will be 80 & 30 .