Question

How do you solve the system of equation -8x-2y=-1 and 3x-12y=4?

Answer 1

Hint: We will solve this set of equations using elimination method. Firstly, we will take an equation, that is, -8x-2y=-1 and multiply this expression with 6 so that y- component in both the equations have the same magnitude and which on subtracting gets cancelled. And we proceed to solve to find the value of \[x\]. We will substitute the value of \[x\], in the equation 3x-12y=4or -8x-2y=-1, and we get the value of \[y\] on solving.

Complete step by step solution:
According to the given question, we have been given a set of two equations having two variables and we have to solve for \[x\] and \[y\]. And we will be using the elimination method to solve the system of equations.
-8x-2y=-1-----(1)
3x-12y=4-----(2)
Since, we can see the y terms in both the equations are different in magnitude, so we will multiply equation (1) by 6 and we get,
\[\left. -8x-2y=-1 \right\}\times 6\]
\[\Rightarrow -48x-12y=-6\]-----(3)
Adding equations (3) - (2), we get,
\[-48x-12y=-6\]
\[\underline{-(3x-12y=4)}\]
\[-51x=-10\]---(4)
Solving the equation (4), we get,
\[\Rightarrow x=\dfrac{10}{51}\]
Putting this value of \[x\] in equation (1), we get,
-8x-2y=-1
\[\Rightarrow -8\left( \dfrac{10}{51} \right)-2y=-1\]
Solving the expression, we have,
\[\Rightarrow -2y=-1+8\left( \dfrac{10}{51} \right)\]
\[\Rightarrow -2y=\left( \dfrac{-51+80}{51} \right)\]
\[\Rightarrow -2y=\left( \dfrac{29}{51} \right)\]
\[\Rightarrow y=\dfrac{-29}{102}\]

Therefore, \[x=\dfrac{10}{51}\And y=-\dfrac{29}{102}\].

Note: The above system of equation can also be solved using substitution method, which is as follows,
Firstly, we will take the equation -8x-2y=-1 and write this equation in terms of \[x\], we get,
-8x-2y=-1
\[\Rightarrow -8x=-1+2y\]
\[\Rightarrow -x=\dfrac{-1+2y}{8}\]
\[\Rightarrow x=\dfrac{1-2y}{8}\]
So we name, \[x=\dfrac{1-2y}{8}\]-----(1)
Now, we will substitute this expression in terms of \[x\] in the equation 3x-12y=4,
We get,
3x-12y=4
\[\Rightarrow 3\left( \dfrac{1-2y}{8} \right)-12y=4\]
Opening up the parenthesis and multiplying the terms, we get,
\[\Rightarrow \left( \dfrac{3(1)-3(2y)}{8} \right)-12y=4\]
Multiplying the terms, we have,
\[\Rightarrow \left( \dfrac{3-6y}{8} \right)-12y=4\]


Taking the LCM in the LHS and then cross multiplying, we get,
\[\Rightarrow \left( \dfrac{3-6y}{8} \right)-\dfrac{96y}{8}=4\]
\[\Rightarrow 3-6y-96y=4\times 8\]
Multiplying \[4\times 8\] and Separating y terms and the constant terms, we get,
\[\Rightarrow -102y=32-3\]
\[\Rightarrow y=\dfrac{-29}{102}\]

Putting the value of \[y=-\dfrac{29}{102}\] in the equation (1), we get,
\[x=\dfrac{1-2y}{8}\]
Substituting the value of y, we get,
\[\Rightarrow x=\dfrac{1-2\left( \dfrac{-29}{102} \right)}{8}\]
Solving we get,
\[\Rightarrow x=\dfrac{1+2\left( \dfrac{29}{102} \right)}{8}\]
\[\Rightarrow x=\dfrac{\left( \dfrac{102+58}{102} \right)}{8}\]
\[\Rightarrow x=\dfrac{\left( \dfrac{160}{102} \right)}{8}\]
\[\Rightarrow x=\dfrac{10}{51}\]