Question 155953
Hi, Hope I can help
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{{{ 0.3x-0.2y=4 }}}
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{{{ 0.2x+0.5y=(-3) }}}
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If you are trying to find the solution to the two equations, this is the way I do it.
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First, we will multiply each equation by 10(multiply each side of the two equations by 10)(you don't have to, but it will get rid of the decimal point)
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We will multiply
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First equation = {{{ 0.3x-0.2y=4 }}} = {{{ 10(0.3x-0.2y)=(10)4 }}} = {{{ 3x - 2y = 40 }}}
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Second equation = {{{ 0.2x+0.5y=-3 }}} = {{{ 10(0.2x+0.5y)=(10)(-3) }}} = {{{ 2x + 5y = (-30) }}}
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The two new equations {{{ (original-equations)(10) }}} are
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First equation = {{{ 3x - 2y = 40 }}}
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Second equation = {{{ 2x + 5y = (-30) }}}
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Now we can start solving
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First, solve for a letter in each equation(doesn't matter which letter) we will solve for "x"
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First equation = {{{ 3x - 2y = 40 }}}
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We will move (-2y) to the right side
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{{{ 3x - 2y + 2y = 40 + 2y }}} = {{{ 3x = 40 + 2y }}}
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We will rearrange the right side
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{{{ 3x = 40 + 2y }}} = {{{ 3x = 2y + 40 }}}
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We will divide by "3" to get "x"
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{{{ 3x/3 = (2y + 40)/3 }}} = {{{ x = (2y + 40)/3 }}}
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Our first answer = {{{ (2y + 40)/3 }}}
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Now we will solve our second equation {{{ 2x + 5y = (-30) }}}
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We will move "5y" to the right side
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{{{ 2x + 5y - 5y = (-30) - 5y }}} = {{{ 2x = (-30) - 5y }}}
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If we rearrange we will get {{{ 2x = (-5y) - 30   }}}
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We need to divide by "2" to get "x"
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{{{ 2x/2 = ((-5y) - 30)/2   }}} = {{{ x = ((-5y) - 30)/2 }}}
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Our second answer = {{{ ((-5y) - 30)/2 }}}
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Our two answers we got are
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Answer 1 =  {{{ (2y + 40)/3 }}}
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Answer 2 = {{{ ((-5y) - 30)/2 }}}
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We will now put the two answers together in an equation (since they equal the same letter "x", they equal each other)
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{{{ (2y + 40)/3 = ((-5y) - 30)/2 }}}
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Now we will start solving for "y", we will cross multiply
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{{{ highlight(2y + 40)/3 = ((-5y) - 30)/highlight(2) }}}
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{{{ (2y + 40)/highlight(3) = highlight((-5y) - 30)/2 }}}
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we will get {{{ 2(2y + 40) = 3((-5y) - 30) }}}
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when we use distribution, we get {{{ 4y + 80 = (-15y) - 90 }}}
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We will move (-15y) to the left
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{{{ 4y + 15y + 80 = (-15y) + 15y - 90 }}} = {{{ 19y + 80 = (-90) }}}
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{{{ 19y + 80 = (-90) }}}
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We will move the "80" to the right
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{{{ 19y + 80 - 80 = (-90) - 80 }}} = {{{ 19y = (-170) }}}
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We will now divide each side by "19" to get "y"
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{{{ 19y/19 = (-170)/19 }}} = {{{ y = (-170)/19 }}}
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We found that "y" = {{{ (-170)/19 }}}
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We can replace "y" with {{{ (-170)/19 }}} in one of our easier equations( one of the {{{ (original - equations)(10) }}} equations
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{{{ 3x - 2y = 40 }}}
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{{{ 2x + 5y = (-30) }}}
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We will use the first equation(replace "y" with {{{ (-170)/19 }}}
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{{{ 3x - 2y = 40 }}}
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{{{ 3x - 2(-170/19) = 40 }}}
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{{{ 3x - (-340/19) = 40 }}}
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double negative
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{{{ 3x + (340/19) = 40 }}}
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we will move {{{ (340/19) }}} to the right side
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{{{ 3x + (340/19) - (340/19) = 40 - (340/19) }}} = {{{ 3x = 40 - (340/19) }}}
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we need to change the "40" , 40 = {{{ 760/19 }}}, replace "40" with that number
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{{{ 3x = (760/19) - (340/19) }}}
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{{{ 3x = (760 - 340)/19) }}} = {{{ 3x = 420/19 }}}
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No we will divide each side by "3" to get our answer
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{{{ 3x/3 = (420/19)/3 }}} = {{{ x = 420/57 }}} = {{{ x = 140/19 }}}
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Our "x" = {{{ 140/19 }}}, we can check our solution in one of our original equations(our very first two equations)
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Equation 1 = {{{ 0.3x-0.2y=4 }}}
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Equation 2 = {{{ 0.2x+0.5y=(-3) }}}
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"x" = {{{ 140/19 }}}
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"y" = {{{ (-170)/19 }}}
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We will replace "x" and "y"
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Equation 1 = {{{ 0.3(140/19) - 0.2((-170)/19) =4 }}}( 0.3 same as {{{(3/10) }}})(0.2 same as {{{ (2/10) }}}
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{{{ (420/190) + (340/190) =4 }}} = {{{ 760/190 =4 }}}
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{{{ 4 =4 }}} (True)
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We can replace "x" and "y" in our second equation ( 0.2 = {{{2/10 }}}, 0.5 = {{{ 5/10 }}} = {{{ 1/2 }}} ) ( we will use {{{ 5/10 }}} )
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{{{ 0.2x+0.5y=(-3) }}}
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{{{ 0.2(140/19)+0.5(-170/19)=(-3) }}}
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{{{ (280/190)+ ((-850)/190)=(-3) }}}
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{{{ (280/190) - ((850)/190)=(-3) }}}
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{{{ (-570)/190= (-3) }}} = {{{ (-3) = (-3) }}} True
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"x" = {{{ 140/19 }}} or 7 {{{ 7/19 }}}
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"y" = {{{ (-170)/19 }}} or ( -8 {{{ 18/19 }}} )
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solution set = (x,y), our solution set = ({{{ 140/19 }}}, {{{ (-170)/19 }}} )
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Hope I helped,Levi