Question

How do you solve the system of equation -8x+3y=7 and 13x-3y=-17?

Answer 1

Hint: We will solve this set of equations using substitution method. We will first take an equation, that is, -8x+3y=7 and rewrite this expression in terms of \[x\]. We will substitute this expression in terms of \[x\], in the equation 13x-3y=-17, we get the value of \[y\] on solving. And when this value of \[y\] is substituted in the equation we earlier wrote in terms of \[x\], will get us the value of \[x\].

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 substitution method to solve the system of equations.
Firstly, we will take the equation -8x+3y=7 and write this equation in terms of \[x\], we get,
-8x+3y=7
\[\Rightarrow 3y=7+8x\]
\[\Rightarrow 8x=3y-7\]
\[\Rightarrow x=\dfrac{3y-7}{8}\]-----(1)
Now, we will substitute this expression in terms of \[x\] in the equation 13x-3y=-17,
We get,
\[13x-3y=-17\]
\[\Rightarrow 13\left( \dfrac{3y-7}{8} \right)-3y=-17\]
Opening up the parenthesis and multiplying the terms, we get,
\[\Rightarrow \left( \dfrac{13(3y)-13(7)}{8} \right)-3y=-17\]
Multiplying the terms, we have,
\[\Rightarrow \left( \dfrac{39y-91}{8} \right)-3y=-17\]
Taking the LCM in the LHS and then cross multiplying, we get,
\[\Rightarrow \dfrac{39y-91}{8}-\dfrac{24y}{8}=-17\]
\[\Rightarrow 39y-91-24y=-17\times 8\]
Multiplying \[-17\times 8\] and Separating y terms and the constant terms, we get,
\[\Rightarrow 39y-24y=-136+91\]
\[\Rightarrow 15y=-45\]
\[\Rightarrow y=-3\]
Putting the value of \[y=-3\] in the equation (1),
\[x=\dfrac{3y-7}{8}\]
Substituting the value of y, we get,
\[\Rightarrow x=\dfrac{3(-3)-7}{8}\]
\[\Rightarrow x=\dfrac{-9-7}{8}\]
Solving we get,
\[\Rightarrow x=\dfrac{-16}{8}=-2\]

Therefore, \[x=-2\And y=-3\].

Note: The above system of equation can also be solved using elimination method, which is as follows,
-8x+3y=7-----(1)
13x-3y=-17-----(2)
Since, we can see the y terms in both the equations are same in magnitude, so don’t need to multiply any factor to it.
Adding equations (1) and (2), we get,
\[-8x+3y=7\]
\[\underline{+(13x-3y=-17)}\]
\[5x=-10\]---(3)
Solving the equation (3), we get,
\[\Rightarrow x=-2\]
Putting this value of \[x\] in equation (1), we get,
-8x+3y=7
\[\Rightarrow -8(-2)+3y=7\]
\[\Rightarrow -8(-2)+3y=7\]
\[\Rightarrow 3y=7-16\]
Solving the expression, we have,
\[\Rightarrow 3y=-9\]
\[\Rightarrow y=-3\]
Therefore, \[x=-2\And y=-3\].