I need help with an algebra assignment

label Algebra
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schedule 1 Day
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Nov 18th, 2015

First Question.

First, we need to add both equations. By this way, the x's will cancel out and then we solve it for y like this.

- x - 5y = 5

x - 5y = 5

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- x + x - 5y - 5y = 5 + 5   ------------>  0 - 10y = 10   -----------> - 10y = 10   -----------> -10y/-10 = 10/-10

y = - 1

Now we enter the obtained y value into the first or second equation and we solve for x like this.

x - 5y = 5   -------------->  x - 5(- 1) = 5   -------------->   x + 5 = 5   --------------> x + 5 - 5 = 5 - 5  -------------> x = 0

So then we write the ordered pair (x , y) which is:  (0 , 5).

The system has a unique solution which is: (x , y) = (0 , 5).

Second Question.

Again, we add both equations. We add what we have on left side of each equation equal to the addition of the expressions that we have on the right side of each equation.

- x + 2y + 8 = 0

x - 2y = 8

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- x + x + 2y - 2y + 8 = 0 + 8   ------------------>  0 + 0 + 8 = 8   ----------------> 8 = 8

So as you can see the x's and y's cancel out at the same time and we get a true equality (8 = 8). Then when this happens it means that we have infinitely many solutions. So finally, we just need to chose the first or the second equation and we solve for y. Then we have:

- x + 2y + 8 = 0  -----------------> - x + x + 2y + 8 = 0 + x  -------------------> 2y + 8 = x

2y + 8 - 8 = x - 8   -------------->  2y = x - 8   -----------------> 2y/2 = x/2 - 8/2  ----------------> y = 1/2x - 4

The system has infinitely many solutions.

They must satisfy the following equation:  y = 1/2x - 4

Nov 18th, 2015

Great  try 1st answer is correct, second answer was not should of been y=x/2 -4

Nov 18th, 2015

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Nov 18th, 2015
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Nov 18th, 2015
Nov 22nd, 2017
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