Question

-3x + 4y = -8
8x - y = 16
Solving systems by substitution​

Answer 1

$\huge{\underline{\underline{\mathcal\color{fuchsia}{Answer}}}}$

The solution to the system of equations is$\implies$$\displaystyle{\red{ \big(\frac{56}{29}, -\frac{16}{29}\big)}}$

Step-by-step explanation:

We are given a system of equations:

$\bold{↬{ }}$$\displaystyle\left \{ {{-3x+4y=-8} \atop {8x-y=16}} \right.$

$\hookrightarrow$ We need to solve these by substitution, so we need to solve one equation for a variable and then substitute the value of that variable into the other equation.

$\hookrightarrow$ After doing this and solving for the opposite variable, we need to insert this into the original equation and solve for the initial variable.

$\hookrightarrow$ Therefore, to solve one of the equations, we will solve it to put it in slope-intercept form and solve for y. The easier equation to work with is equation two.

$\begin{gathered}\displaystyle 8x - y = 16\\\\-y = -8x + 16\\\\\frac{-y}{-1}=\frac{-8x+16}{-1}\\\\y = 8x - 16\end{gathered}$

$\longrightarrow$ Now, we've solved for y. So, we can substitute this into either equation and solve for x.

$\begin{gathered}\displaystyle -3x + 4(8x -16)=-8\\\\-3x + 32x - 64 = -8\\\\29x - 64 = -8\\\\29x = 56\\\\\frac{29x}{29}=\frac{56}{29}\\\\x = \frac{56}{29}\end{gathered}$

$\rightsquigarrow$ Now, we substitute our value for x into one of the original equations and solve for y.

$\begin{gathered}\displaystyle 8\big(\frac{56}{29}\big)-y=16\\\\-y = 16 - 8\big(\frac{56}{29}\big)\\\\-y=\frac{16}{29}\\\\\frac{-y}{-1}=\frac{\frac{16}{29}}{-1}\\\\y = -\frac{16}{29}\end{gathered}$

$\rightsquigarrow$ Therefore, the solution to our system of equations is:

$\displaystyle \big(\frac{56}{29}, -\frac{16}{29}\big)$

Answer 2

Answer :

x = 56/29 , y = -16/29

Solution :

Here ,

The given system of linear equations is ;

-3x + 4y = -8 --------(1)

8x - y = 16 --------(2)

We need to solve the given system by substitution method .

Now ,

Considering eq-(2) , we have ;

=> 8x - y = 16

=> y = 8x - 16 ------(3)

Now ,

Putting y = 8x - 16 in eq-(1) , we get ;

=> -3x + 4y = -8

=> -3x + 4(8x - 16) = -8

=> -3x + 32x - 64 = -8

=> 32x - 3x = 64 - 8

=> 29x = 56

=> x = 56/29

Now ,

Putting x = 56/29 in eq-(3) , we get ;

=> y = 8x - 16

=> y = 8(56/29) - 16

=> y = 448/29 - 16

=> y = (448 - 464)/29

=> y = -16/29

Hence ,

x = 56/29 , y = -16/29