Question

Solve the linear equasion

y=-2x-8 and y=6x+1

Answer 1

what exactly are u looking for

Answer 2

I guess you are trying to find the value of x in which both the equations will be the same or intercept?
you can make the equations equal to each other
\[-2x-8=6x+1\]
\[-2x-6x=1+8\]
\[-8x=9\]
\[x=\frac{ - 9}8{ ? }\]

Answer 3

Write the solution as an ordered pair like this: (10, -7)

Answer 4

First put the equations in standard form:
A) 2x + y = -8
B) -6x +y = +1
Multiply equation B by minus 1
B) 6x -y = -1 then add this to A)
A) 2x + y = -8
8x = -9
x = -1.125
Inputting this value into equation A
2*-1.125 +y = -8
-2.25 +y = -8
y = -5.75

Answer 5

I think you mean "solve the system of linear equations."

y = -2x - 8
y = 6x + 1

Subtract the second equation from the first equation:

0 = -8x - 9

Add 8x to both sides

8x = -9

Divide both sides by 8

\(x = -\dfrac{9}{8} \)

Now that we know the value of x, substitute x in the first equation with the klnown value.

y = -2x - 8

\(y = -2 \cdot \left( -\dfrac{9}{8} \right) - 8 \)

Write 2 as a fraction to multiply

\(y = -\dfrac{2}{1} \cdot \left( -\dfrac{9}{8} \right) - 8 \)

Multiply the fractions

\(y = \dfrac{18}{8} - 8 \)

Reduce the fraction

\(y = \dfrac{9}{4} - 8 \)

Write 8 as a fraction with denominator 4 to be able to add

\(y = \dfrac{9}{4} - \dfrac{32}{4} \)

Add fractions

\(y = -\dfrac{23}{4}\)

The solution is

\( \left( -\dfrac{9}{8}, -\dfrac{23}{4} \right) \)