Question 156745
Hi, Hope I can help,
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2x-2y=15
4x+4y=10
Help, I need to solve for x and y
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First, you can reduce the second equation, by multiplying each side by "2"
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{{{4x+4y=10}}} = {{{(4x+4y)/2=10/2}}} = {{{(4x+4y)/2=10/2}}} = {{{(4x)/2+(4y)/2=10/2}}} = {{{2x+2y=5}}}
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The second equation {{{ 4x+4y=10 }}} is reduced to {{{ 2x+2y=5 }}}
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The Two equations are
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{{{ 2x-2y=15 }}}
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{{{ 2x+2y=5 }}}
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This is the way I usually solve these equations, First, solve for a letter(usually the easiest letter), we will solve for "x" in both equations
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Equation 1 = {{{ 2x-2y=15 }}}
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Now we will move (-2y) to the right side
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{{{ 2x-2y=15 }}} = {{{ 2x-2y + 2y=15 + 2y }}} = {{{ 2x = 15 + 2y }}}
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We will rearrange the numbers
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{{{ 2x = 15 + 2y }}} = {{{ 2x = 2y + 15 }}}
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We will divide each side by "2" to get "x"
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{{{ 2x/2 = (2y + 15)/2 }}} = {{{ x = (2y + 15)/2 }}}
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Our first "x" = {{{ (2y + 15)/2 }}}
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Now let's solve "x" in our second equation, {{{ 2x+2y=5 }}}
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We will move (2y) to the right side
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{{{ 2x+2y=5 }}} = {{{ 2x+2y - 2y=5 -2y }}} = {{{ 2x= 5 -2y }}}
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We will rearrange the numbers
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{{{ 2x= 5 -2y }}} = {{{ 2x= (-2y) + 5 }}}
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We will divide each side by "2" to get "x"
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{{{ 2x= (-2y) + 5 }}} = {{{ 2x/2= ((-2y) + 5)/2 }}} = {{{ x= ((-2y) + 5)/2 }}}
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Our second "x" = {{{ ((-2y) + 5)/2 }}}
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Since "x" is one number, and "x" equals our two answers, our answers equal each other, we can put our two answers in an equation
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Answer 1 = {{{ (2y + 15)/2 }}}
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Answer 2 = {{{ ((-2y) + 5)/2 }}}
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{{{ (2y+15)/2 = ((-2y) + 5)/2 }}}
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We will now solve for "y"
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{{{ (2y+15)/2 = ((-2y) + 5)/2 }}}
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We will use cross multiplication
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{{{ highlight(2y+15)/2 = ((-2y) + 5)/highlight(2) }}} = {{{ (2y+15)/highlight(2) = highlight((-2y) + 5)/2 }}}
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It will become {{{ (2y+15)(2) = (2)((-2y)+5) }}}
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We will rearrange numbers
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{{{ (2y+15)(2) = (2)((-2y)+5) }}} = {{{ (2)(2y+15) = (2)((-2y)+5) }}}
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We will use the distribution method
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{{{ highlight(2)(highlight(2y)+15) = highlight(2)(highlight(-2y)+5) }}} =
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{{{ highlight(2)(2y+highlight(15)) = highlight(2)((-2y)+highlight(5)) }}} =
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{{{ 4y + 30 = (-4y) + 10 }}}
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Now we will solve for "y", we will move (-4y) to the left
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{{{ 4y + 30 = (-4y) + 10 }}} = {{{ 4y + 4y + 30 = (-4y) + 4y + 10 }}} = {{{ 8y + 30 = 10 }}}
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Now we will move (30) to the right side
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{{{ 8y + 30 = 10 }}} = {{{ 8y + 30 -30 = 10 -30 }}} = {{{ 8y = (-20) }}}
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Now to find "y" we will divide each side by "8"
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{{{ 8y = (-20) }}} = {{{ 8y/8 = (-20)/8 }}} = {{{ y = (-20)/8 }}}, if we reduce {{{ y = (-5)/2 }}}
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y = {{{ (-5)/2 }}}
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To find "x" all you do is substitute "y" with {{{ (-5)/2 }}} in one of our original equations
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{{{ 2x-2y=15 }}}
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{{{ 2x+2y=5 }}}
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Let's use the second equation(replace "y" with {{{ (-5)/2 }}} )
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{{{ 2x+2y=5 }}} = {{{ 2x+2((-5)/2)=5 }}} = {{{ 2x+(-5)=5 }}} = {{{ 2x-5 =5 }}}
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We will move (-5) over to the right side
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{{{ 2x-5 =5 }}} = {{{ 2x-5 + 5 =5 + 5 }}} = {{{ 2x =10 }}}
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Now we will divide each side by "2" to get "x"
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{{{ 2x/2 =10/2 }}} = {{{ x =10/2 }}} = {{{ x =5 }}}
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x = {{{ 5 }}}
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We can check our answers by replacing "x" and "y" with our answers we got, in our very first equations
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x = {{{ 5 }}}
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y = {{{ (-5)/2 }}}
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Let's replace "x" and "y" with the numbers above
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{{{ 2x-2y=15 }}} = {{{ 2(5)-2((-5)/2))=15 }}} = {{{ 10 -(-5)=15 }}} = {{{ 10 +5 =15 }}} = {{{ 15 =15 }}} (True)
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{{{ 4x+4y=10 }}} = {{{ 4(5)+4((-5)/2)=10 }}} = {{{ 20+(-10)=10 }}} = {{{ 20-10=10 }}} = {{{ 10=10 }}} (True)
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This method will work for all problems like this
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x = {{{ 5 }}}
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y = {{{ (-5)/2 }}}
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The ordered pair is (x,y), our ordered pair is (5, {{{ (-5)/2 }}} )
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Hope I helped, Levi