Solveeit Logo
Login

Question

Question: How do you solve the system of linear equations \[ - 8x + 3y = 7\] and \[13x - 3y = - 1\] ?...

How do you solve the system of linear equations −8x+3y=7 - 8x + 3y = 7 and 13x−3y=−113x - 3y = - 1 ?

Explanation

Solution

We use the substitution method to solve two linear equations given in the question. We find the value of 3y from the first equation in terms of x and substitute in the second equation which becomes an equation in x entirely. Solve for the value of x and substitute back the value of y to obtain the value of y.

Complete step-by-step answer:
We have two linear equations −8x+3y=7 - 8x + 3y = 7 and 13x−3y=−113x - 3y = - 1
Let us solve the first equation to obtain a value of 3y in terms of x.
We have −8x+3y=7 - 8x + 3y = 7
Shift the value of x to RHS of the equation.
⇒3y=7+8x\Rightarrow 3y = 7 + 8x … (1)
Now we substitute the value of 3y=8x+73y = 8x + 7from equation (1) in the second linear equation.
Substitute 3y=8x+73y = 8x + 7in 13x−3y=−113x - 3y = - 1
⇒13x−(8x+7)=−1\Rightarrow 13x - (8x + 7) = - 1
Open the bracket in LHS of the equation
⇒13x−8x−7=−1\Rightarrow 13x - 8x - 7 = - 1
Add like terms in numerator of LHS of the equation
⇒5x−7=−1\Rightarrow 5x - 7 = - 1
Shift the constant values to one side of the equation.
⇒5x=−1+7\Rightarrow 5x = - 1 + 7
⇒5x=6\Rightarrow 5x = 6
Divide both sides by 5
⇒5x5=65\Rightarrow \dfrac{{5x}}{5} = \dfrac{6}{5}
Cancel same terms from numerator and denominator.
⇒x=65\Rightarrow x = \dfrac{6}{5}
Now substitute the value of x=65x = \dfrac{6}{5}in equation (1) to get the value of y
⇒3y=7+8×(65)\Rightarrow 3y = 7 + 8 \times \left( {\dfrac{6}{5}} \right)
Open the bracket in RHS of the equation.
⇒3y=7+485\Rightarrow 3y = 7 + \dfrac{{48}}{5}
Take LCM in RHS of the equation
⇒3y=35+485\Rightarrow 3y = \dfrac{{35 + 48}}{5}
⇒3y=835\Rightarrow 3y = \dfrac{{83}}{5}
Cross multiply 3 from numerator of LHS to RHS
⇒y=8315\Rightarrow y = \dfrac{{83}}{{15}}

∴\therefore Solution of the system of linear equations is x=65;y=8315x = \dfrac{6}{5};y = \dfrac{{83}}{{15}}

Note:
We can use combination method to solve the system of linear equations as there are exact same values in both linear equations with opposite signs.
We have equations −8x+3y=7 - 8x + 3y = 7 and 13x−3y=−113x - 3y = - 1
Multiply second equation by -1
Then equations become −8x+3y=7 - 8x + 3y = 7 and −13x+3y=1 - 13x + 3y = 1
Now we subtract second equation from first equation

\- 8x + 3y = 7 \\\ \underline { - 13x + 3y = 1} \\\ 5x = 6 \\\ \end{gathered} $$ Cross multiply 5 from LHS to RHS $$ \Rightarrow x = \dfrac{6}{5}$$ Now substitute the value of $$x = \dfrac{6}{5}$$in equation (1) to get the value of y $$ \Rightarrow 3y = 7 + 8 \times \left( {\dfrac{6}{5}} \right)$$ Open the bracket in RHS of the equation. $$ \Rightarrow 3y = 7 + \dfrac{{48}}{5}$$ Take LCM in RHS of the equation $$ \Rightarrow 3y = \dfrac{{35 + 48}}{5}$$ $$ \Rightarrow 3y = \dfrac{{83}}{5}$$ Cross multiply 3 from numerator of LHS to RHS $$ \Rightarrow y = \dfrac{{83}}{{15}}$$ $$\therefore $$Solution of the system of linear equations is $$x = \dfrac{6}{5};y = \dfrac{{83}}{{15}}$$