Question

Solved by substitution for y=x-6y=8x+1​

Answer 1

Here is your system of equations:

$\left \{ {{y=x-6} \atop {y=8x+1}} \right.$

When solving a system of equations, you can either eliminate or substitute, however, I will be substituting since that's what the question is asking.

In the first equation, if y is equal to x - 6, I can substitute it with the y in the second equation with 8x + 1.

Here is your new equation:

$x-6 = 8x+1$

Since you have variables on both sides, you have to cancel it on either side. In the equation on the left, the variable is x. To cancel it out, you have to do the opposite of it, which is -x, or negative x. That means you have to subtract x to both equations.

$x-6-x = 8x+1-x$

$-6=7x+1 \rightarrow 7x+1=-6$

Now you have one variable and two regular integers. You have to leave the variable x alone to find its value. Along with it is +1, or positive 1. To remove it, you have to do the opposite of it, which is -1 or subtract 1.

$7x+1-1=-6-1$

$7x=-7$

• Remember that when you subtract from a negative number, the number will increase but the sign will stay negative.

Now you need x to stay alone, so divide both sides by 7.

$\frac{7x}{7} =\frac{-7}{7} \\ \downarrow \\ x=-1$

Remember:

NEGATIVES

• Negative(-) times(×) negative(-) = positive(+)
• Negative(-) times(×) positive(+) = negative(-)
• Negative(-) divided(÷) by positive(+) = negative(-)
• Negative(-) divided(÷) by negative(-) = positive(+)

POSITIVES

• Positive(+) times(×) positive(+) = positive(+)
• Positive(-) times(×) negative(-) = negative(-)
• Positive(+) divided(÷) by positive(+) = positive(+)
• Positive(+) divided(÷) by negative(-) = negative(-)

Now that you have the value of x, you can substitute it with the x in one of the equations.

$y=x-6 \rightarrow y=(-1)-6$

Subtract -1 by 6:

$-1-6=-7$

That means y is equal to -7.

$\left \{ {{x=-1} \atop {y=-7}} \right.$

Your answer is (-1, -7).

Answer 2

Answer:

y=y

x-6=8x+1

-7x=7

x=-1

y=-7