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How do you use the substitution method to solve #y=6x+5, 6x-y=7#?

Answer 1

Nathan Axel

This system has no solutions. See explanation.

The variable #y# is calculated in the first equation, so we can substitute it to the second:

#6x-(6x+5)=7#

#6x-6x-5=7#

#-5=7#

This equation is false, so the starting system has no solutions.

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Answer 2

Jace Anderson

To solve the system of equations using the substitution method:

Solve one of the equations for one variable in terms of the other.
Substitute the expression found in step 1 into the other equation.
Solve the resulting equation for the remaining variable.
Once you have found the value of one variable, substitute it back into one of the original equations to find the value of the other variable.

Given the system:

y = 6x + 5
6x - y = 7

From equation 1, solve for y: y = 6x + 5

Now, substitute the expression for y into equation 2: 6x - (6x + 5) = 7

Now, solve for x: 6x - 6x - 5 = 7 -5 = 7

Since -5 does not equal 7, there is no solution to the system of equations.

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.