How do you solve the system #2x+7y=3 # and #x = 1-4y# by substitution?

Answer 1
Substitute the formula for #x# from the second equation into the first equation as follows:
#3 = 2x+7y = 2(1-4y)+7y = 2-8y+7y = 2-y#
Add #y# to both ends to get:
#y + 3 = 2#

Subtract #3 from both sides to get:

#y = -1#
Substitute this value for #y# into the original second equation to find #x#:
#x = 1 - 4y = 1 - 4(-1) = 1+4 = 5#
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Answer 2

Try this:

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Answer 3

To solve the system using substitution, substitute the value of x from the second equation into the first equation.

From x = 1 - 4y, substitute x = 1 - 4y into the equation 2x + 7y = 3:

2(1 - 4y) + 7y = 3

Now, solve for y:

2 - 8y + 7y = 3

  • y = 1 y = -1

Next, substitute y = -1 into x = 1 - 4y:

x = 1 - 4(-1) x = 1 + 4 x = 5

Therefore, the solution to the system is x = 5 and y = -1.

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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.

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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