Question 160339
Hi, Hope I can help,
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The area of a rectangle is 30 cm^2. The perimeter is 26cm. What are the length and width of the rectangle. 
this is what i have so far
a=lw so 30= lw 
p=2L + 2W so 26= 2L+2W 
I'm not sure if I am supposed to substitute the first equation into the second one or just try to isolate L and w. 
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You did everything right so far.
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You have the two equations
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{{{ LW = 30 }}}
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{{{ 2L+2W = 26 }}}
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 (You can reduce the second equation by dividing each side by "2", {{{ 2L+2W = 26 }}} = {{{ (2L+2W)/2 = 26/2 }}} = {{{ (2L/2)+(2W/2) = 13 }}} = {{{ L+W = 13 }}} )
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The two equations will =
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{{{ LW = 30 }}}
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{{{ L+W = 13 }}}
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We have two equations, or a system of equations
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To find the "L" and "W", we will use the way I solve systems of equations, which is pretty easy
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{{{ LW = 30 }}}
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{{{ L+W = 13 }}}
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First solve for a letter ( doesn't matter which one, usually the easiest), we will solve for "W" in both equations
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First equation {{{ LW = 30 }}}
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To get "W" by itself we will divide each side by "L"
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{{{ LW = 30 }}} = {{{ LW/L = 30/L }}} = {{{ W = 30/L }}}, Our first answer = {{{ 30/L }}}
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Second equation {{{ L+W = 13 }}}
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To get "W" by itself we will move "L" to the right side
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{{{ L+W = 13 }}} = {{{ L-L+W = 13-L }}} = {{{ W = 13-L }}} or {{{ W = (-L) + 13 }}}
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Our second answer = {{{ (-L) + 13 }}}
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We can put our two answers in an equation, since both answers represent "W"(width), our two answers equal each other
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{{{ 30/L = (-L) + 13 }}} = {{{ 30/L = ((-L) + 13)/1 }}}
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We will use cross multiplication
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{{{ 30/L = ((-L) + 13)/1 }}} = {{{ highlight(30)/L = ((-L) + 13)/highlight(1) }}} = {{{ 30/highlight(L) = highlight((-L) + 13)/1 }}}
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It becomes {{{ 30 = -(L^2)+ 13L }}} = {{{ -(L^2)+ 13L  = 30 }}}
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This is a quadratic equation, we will put it in standard form(move "30" to the left side)
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{{{ -(L^2)+ 13L  = 30 }}} = {{{ -(L^2)+ 13L -30 = 30-30 }}} = {{{ -(L^2)+ 13L -30 = 0 }}}
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Standard form {{{ ax^2+bx+c=0 }}} Since this is a quadratic equation, we can use quadratic formula( We can solve by factoring, or completing the square as well, but I believe the formula is easiest)
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{{{ ax^2+bx+c=0 }}}, {{{ -(L^2)+ 13L -30 = 0 }}}, (a = (-1), b = 13, c = (-30)

quadratic formula = {{{ L = (-b +- sqrt( b^2-4*a*c ))/(2*a) }}}, replace the letters with numbers
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{{{ L = (-(13) +- sqrt( (13)^2-4*(-1)*(-30) ))/(2*(-1)) }}} = {{{ L = (-13 +- sqrt( 169-120 ))/(-2)) }}} = {{{ L = (-13 +- sqrt( 49))/(-2)) }}} = {{{ L = (-13 +- 7)/(-2)) }}} =
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{{{ L = (-13 + 7)/(-2)) }}} = {{{ x = (-6)/(-2)) }}} = {{{ x = 3 }}}
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{{{ L = (-13 - 7)/(-2)) }}} = {{{ x = (-20)/(-2)) }}} = {{{ x = 10 }}}
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"L" can be either {{{ 10 }}} or {{{ 3 }}}
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Just replace "L" in the second equation
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{{{ L+W = 13 }}} = {{{ 10+W = 13 }}} = {{{ W = 3 }}}, or 
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{{{ L+W = 13 }}} = {{{ 3+W = 13 }}} = {{{ W = 10 }}}
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Most likely {{{ L = 10 }}} and {{{ W = 3 }}}, it could be the other way around, you can check by replacing these answers in the first two original equations.
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Hope I helped, Levi