SOLUTION: 4-(y-3) = x+3 1st equation 10+x ―—―— = 2+(y-12) 2nd equation 3

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Question 148873: 4-(y-3) = x+3 1st equation
10+x
―—―— = 2+(y-12) 2nd equation
3

Answer by Electrified_Levi(103) About Me  (Show Source):
You can put this solution on YOUR website!
Hi, Hope I can help,
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+4-%28y-3%29+=+x%2B3+ (1st equation)
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+%2810%2Bx%29%2F3+=+2%2B%28y-12%29+ (2nd equation)
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If you are trying to solve for "x" and "y", this is how you do it.
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First, you have to reduce each equation to (Ax+By=C) form, we will do the first equation.
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+4-%28y-3%29+=+x%2B3+
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We will take the parentheses off
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+4+-+y+%2B+3+=+x+%2B+3+
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We will add on the left side
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+4+%2B+3+-+y+=+x+%2B+3+
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+7+-+y+=+x+%2B+3+
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We will move the "y" to the right side
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+7+-+y+%2B+y+=+x+%2B+3+%2B+y+
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+7+=+x+%2B+y+%2B+3+
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We will now move the "3" to the left side
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+7+-+3+=+x+%2B+y+%2B+3+-+3+
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+4+=+x+%2B+y+
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Our first equation is (x + y = 4)
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Now we will solve the second equation
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+%2810%2Bx%29%2F3+=+2%2B%28y-12%29+
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We can multiply everything by "3" to get rid of the fraction
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+%28%2810%2Bx%29%2F3%29%283%29+=+2%283%29%2B%28y-12%29%283%29+
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It will get rid of the denominator of the fraction
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+10%2Bx+=+6+%2B+3%28y-12%29+
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We will get rid of the parentheses on the right side, using distribution method
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+10%2Bx+=+6+%2B+3y+-+36+
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We will subtract the right side
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+10%2Bx+=+6+-+36+%2B+3y++
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+10%2Bx+=+%28-30%29+%2B+3y++
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We will move "3y" to the left side
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+10+%2B+x+-+3y+=+%28-30%29+%2B+3y+-+3y++
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+10+%2B+x+-+3y+=+%28-30%29+
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We will move "10" to the right side
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+10+-+10+%2B+x+-+3y+=+%28-30%29+-+10+
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+x+-+3y+=+%28-40%29+
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Our reduced second equation is +x+-+3y+=+%28-40%29+
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We can now put the two equations side by side and solve for "x" and "y"
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x + y = 4
x - 3y = (-40)
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I will show you two different ways to solve these problems, one is a method that can be done on certain problems, the other you can do on any problem.
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(First Method) If you can either add or subtract the equations to get rid of a letter, this method works, we can subtract the equations to get rid of "x"
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x + y = 4
- (x - 3y = (-40))
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Getting rid of parentheses
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x + y = 4
-x + 3y = +40
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It will become
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0x + 4y = 44
4y = 44
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We can divide everything by "4" to get "y"
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+4y%2F4+=+44%2F4+
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+y+=+11+
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We can now replace "y" with "11" in one of our equations
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x + y = 4
x + (11) = 4
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We can move "11" to the right side
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x + 11 - 11 = 4 - 11
x = (-7)
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We can check by replacing "x" and "y" in our second equation
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x - 3y = (-40)
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(-7) - 3(11) = (-40)
(-40)=(-40)
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The solution set is (x,y), our solution set is (-7,11)
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(Second Method) This method works on any problem like this,
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First, we solve for a letter(doesn't matter which one, usually the easiest one) in both of the equations, we will use "x"
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(First Equation) x + y = 4
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We will move the "y" to the right side
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x + y - y = 4 - y
(x = 4 - y), or (x = (-y) + 4)
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((-y) + 4) is our first answer
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We will now solve "x" in our second equation
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x - 3y = (-40)
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We will move (-3y) to the right side
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x - 3y + 3y = (-40) + 3y
(x = (-40) + 3y), or (x = 3y - 40)
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(3y - 40) is our second answer
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We can put the two answers together, since they equal each other
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((-y) + 4) = (3y - 40)
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We will get rid of the first sets of parentheses
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(-y) + 4 = 3y - 40
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We will move (-y) to the right side
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(-y) + y + 4 = 3y + y - 40
4 = 4y - 40
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We will move (-40) to the left side
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4 + 40 = 4y - 40 + 40
44 = 4y
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We can divide everything by 4
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+44%2F4+=+4y%2F4+
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11 = y
y = 11
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We will now replace "y" with "11" in our second equation(we will solve "x" a different way this time)
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x - 3y = (-40)
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x - 3(11) = (-40)
x - 33 = (-40)
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We will move (-33) to the right side
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x - 33 + 33 = (-40) + 33
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x = (-7)
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We can replace "x" and "y" in our first equation, to check our answers
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x + y = 4
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(-7) + (11) = 4
4 = 4
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Our answers are
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x = (-7)
y = (11)
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Our solution set is (-7,11)
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Hope I helped, Levi