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

This system has no solutions. See explanation.

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

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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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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.

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.

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.

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