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Part 1:

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$x$:

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$\\frac{(\\var{numh[0]}-\\var{num1p[1]})}{(\\var{num1p[0]}-\\var{num1p[2]})} = \\var{ans11}= \\var{precround({ans11},2)}$ 

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Part 2:

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$x$:

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$\\frac{((\\var{num1p[7]} \\times \\var{num1p[6]})-(\\var{num1p[4]} \\times \\var{num1n[2]})}{((\\var{num1p[1]} \\times \\var{num1p[5]})-(\\var{num1p[7]} \\times \\var{num1n[7]})} = \\var{ans21}= \\var{precround({ans21},2)}$ 

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Part 3:

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$x$:

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$\\frac{((\\var{num3p[4]} \\times \\var{num3p[5]})+\\var{num3p[6]}+(\\var{num3n[4]} \\times \\var{num3n[5]})-\\var{num3p[0]}-(\\var{num3p[1]} \\times \\var{num3n[0]})-(\\var{num3n[1]} \\times \\var{num3n[2]}))}  {((\\var{num3p[1]} \\times \\var{num3p[2]})+(\\var{num3n[1]} \\times \\var{num3p[3]})-(\\var{num3p[4]} \\times \\var{num3n[3]})-(\\var{num3n[4]} \\times \\var{num3p[7]}))} = \\var{ans31}= \\var{precround({ans31},2)}$ 

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Part 4:

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$x$:

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$\\frac{(\\var{nega[3]}-\\frac{\\var{nega[1]}}{\\var{posa[1]}}+\\frac{\\var{posa[5]}}{\\var{posa[3]}}-\\frac{\\var{nega[2]}}{\\var{posa[6]}})}  {(\\frac{\\var{posa[2]}}{\\var{posa[1]}}-\\frac{\\var{posa[4]}}{\\var{posa[3]}}+\\frac{\\var{posa[7]}}{\\var{posa[6]}})} = \\var{ans41}= \\var{precround({ans41},2)}$ 

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Part 5:

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$x$:

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$\\frac{((\\var{posb[3]} \\times \\var{nega[4]}) - (\\var{posb[1]} \\times \\var{posb[5]}))}{((\\var{posb[1]} \\times \\var{posb[4]}) - (\\var{posb[3]} \\times \\var{posb[2]}))} = \\var{ans51}= \\var{precround({ans51},2)}$ 

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$\\var{num1p[0]}x + \\var{num1p[1]} = \\var{num1p[2]}x + \\var{numh[0]}$

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$x =$  [[0]]

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Solving for a variable which is present on both sides requires collecting those terms first and then rearranging in the usual way.

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For example,

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Take the equation:    $2x-5=15-3x

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The terms in $x$ are moved, ideally in a way which results in a positive coefficient:    $2x+3x-5=15\\;\\;\\;\\therefore\\;\\;\\;5x-5=15$

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Finally it is solved for $x$:    $5x=15+5=20\\;\\;\\;\\therefore\\;\\;\\;x=\\frac{20}{5}=4$

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$\\var{num1p[4]}(\\var{num1p[5]}  \\var{num1n[2]}x) = \\var{num1p[7]}(\\var{num1p[6]}  \\var{num1n[7]}x)$

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$x =$  [[0]]

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While there is more than one approach to solving equations of this form, a straightforward step would be to expand the brackets first and then rearrange and simplify according to the process described in Part a).

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$\\var{num3p[0]} + \\var{num3p[1]}(\\var{num3p[2]}x \\var{num3n[0]})\\var{num3n[1]}(\\var{num3p[3]}x \\var{num3n[2]}) = \\var{num3p[4]}(\\var{num3p[5]}  \\var{num3n[3]}x) + \\var{num3p[6]} \\var{num3n[4]}(\\var{num3p[7]}x \\var{num3n[5]})$

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$x =$  [[0]]

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Again, the brackets should be expanded first to allow for simple addition and subtraction of like terms and final simplification.

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$\\frac{1}{\\var{posa[1]}}(\\var{posa[2]}x \\var{nega[1]})-\\frac{1}{\\var{posa[3]}}(\\var{posa[4]}x +\\var{posa[5]})+\\frac{1}{\\var{posa[6]}}(\\var{posa[7]}x \\var{nega[2]}) = \\var{nega[3]}$

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$x =$  [[0]]

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You may expand the brackets and systematically collect like terms as before, but the arithmetic with the fractions may be more involved than the other method outlined below.

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Find a common denominator of the fractional coefficients and convert all fractions into their equivalent forms with this common denominator. You can then mulptiply through all terms by this common denominator and finish the process according to the steps in the previous questions.

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For example,

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$\\frac{1}{3}(4+8x)-\\frac{2}{9}(12-x)+\\frac{3}{2}(5x+7)=5$

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Converting all fractions to equivalent form with common denominator:    $\\frac{6}{18}(4+8x)-\\frac{4}{18}(12-x)+\\frac{27}{18}(5x+7)=5$

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Multiplying all terms through by this common denominator:    $6(4+8x)-4(12-x)+27(5x+7)=90$

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The brackets are then expanded:    $24+48x-48+4x+135x+189=90$

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Terms are collected:    $187x=-75\\;\\;\\;\\therefore\\;\\;\\;x=-\\frac{75}{187}$

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Solve for $x$ in the following, to 2 decimal places:

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Watch the video below for help with the questions:

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Solving Equations

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Simple Linear Equations involving basic transposition.

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