Question

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

Answer 1

Answer:

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

Step-by-step explanation:

We are given a system of equations:

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

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

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

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

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

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

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

Therefore, the solution to our system of equations is:

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