Question

How do you solve by elimination with fraction?

Answer 1

Hint: We are going to explain this by taking an example. Let’s take some linear equation in two variables which contain fractions. In elimination of fraction we eliminate the fractions by multiplying each side of the equation by a common denominator. Now we can solve the resulting system using the addition method or elimination method or substitution method.

Complete step-by-step answer:
Now consider system of equations with two variables,
 \[\dfrac{1}{4}x + \dfrac{1}{2}y = 8{\text{ - - - - - (1)}}\]
and \[\dfrac{1}{8}x + \dfrac{1}{6}y = 4{\text{ - - - - - (2)}}\] .
We need to eliminate the fractions in both equations.
Now consider equation (1),
 \[\dfrac{1}{4}x + \dfrac{1}{2}y = 8\] multiply this equation by ‘4’ on both sides of the equation,
 \[ \Rightarrow 4 \times \dfrac{1}{4}x + 4 \times \dfrac{1}{2}y = 8 \times 4\]
 \[x + 2y = 32{\text{ - - - - - - - (3)}}\]
Now consider equation (2),
 \[\dfrac{1}{8}x + \dfrac{1}{6}y = 4\] multiply this equation by ‘8’ on both sides of the equation,
 \[ \Rightarrow 8 \times \dfrac{1}{8}x + 8 \times \dfrac{1}{6}y = 1 \times 8\]
 \[ \Rightarrow x + \dfrac{4}{3}y = 8\] .
But we still have a fraction in this equation. So we multiply ‘3’ on both sides of the equation,
 \[3x + 3 \times \dfrac{4}{3}y = 3 \times 8\]
 \[3x + 4y = 24{\text{ - - - - - - (4)}}\]
Thus we have a system of equations \[x + 2y = 32\] and \[3x + 4y = 24{\text{ }}\] .
We can solve this by making one of the coefficient of a variable same and subtracting we get the required result,
We have, \[3x + 4y = 24{\text{ - - - - - - (4)}}\]
We multiply 3 to the equation (3) on both sides we have,
 \[3 \times x + 2 \times 3y = 32 \times 3\]
 \[3x + 6y = 96{\text{ - - - - - - - (5)}}\]
Now subtracting equation (5) with equation(4) we have,
 \[ \Rightarrow 3x + 6y - (3x + 4y) = 96 - 24\]
 \[ \Rightarrow 3x + 6y - 3x - 4y = 72\]
Cancelling the ‘x’ variable term, we have
 \[ \Rightarrow 6y - 4y = 72\]
 \[ \Rightarrow 2y = 72\]
Divide by 2 on both sides,
 \[ \Rightarrow y = \dfrac{{72}}{2}\]
 \[ \Rightarrow y = 36\]
We can obtain the value of ‘x’ by substituting in any of the above equations.
Now we substitute \[y = 36\] in equation (3) we have,
 \[ \Rightarrow x + 2 \times 36 = 32\]
 \[ \Rightarrow x + 72 = 32\]
 \[ \Rightarrow x = 32 - 72\]
 \[ \Rightarrow x = - 40\]
Thus we have the variable values, \[x = - 40\] and \[y = 36\]
So, the correct answer is “ \[x = - 40\] and \[y = 36\] ”.

Note: All we did is in above is converting the given system of equation containing fraction into a simple linear system of equation and we solved it. Follow the same procedure for any given pair of linear equations with two variables. We have many methods to solve these equations; those are elimination method, cross multiplication method and substitution method. In all the cases we will get the same answer.