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Linear Equations in One Variable

Short Answer Type

111.

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102 Views

112.

Solve for $straight x colon fraction numerator 7 straight x plus 14 over denominator 3 end fraction minus fraction numerator 17 minus 3 straight x over denominator 5 end fraction equals 6 straight x minus fraction numerator 4 straight x plus 2 over denominator 3 end fraction minus 5$

158 Views

113.

Solve for $space space space space straight x colon fraction numerator straight x minus left parenthesis 7 minus 8 straight x right parenthesis over denominator 9 straight x minus left parenthesis 3 plus 4 straight x right parenthesis end fraction equals 2 over 3$

117 Views

114.
Solve for $straight x colon fraction numerator left parenthesis 2 plus straight x right parenthesis left parenthesis 7 minus straight x right parenthesis over denominator left parenthesis 5 minus straight x right parenthesis left parenthesis 4 plus straight x right parenthesis end fraction equals 1$

We have $fraction numerator left parenthesis 2 plus straight x right parenthesis left parenthesis 7 minus straight x right parenthesis over denominator left parenthesis 5 minus straight x right parenthesis left parenthesis 4 plus straight x right parenthesis end fraction equals 1$

By cross multiplication, we get:

                    (2 + x)(7 - x) = (5 - x)(4 + x)

$rightwards double arrow$              2(7 - x) + x(7 - x) = 5(4 + x) - x(4 + x)

$rightwards double arrow$                14 - 2x + 7x - x²=20 + 5x - 4x -x²

$rightwards double arrow$ -x²+ x²- 2x + 7x - 5x + 4x = 20 - 14

$rightwards double arrow$                  -7x + 7x + 4x = 6

$rightwards double arrow$                            4x = 6

$rightwards double arrow$ x = $6 over 4$ or $3 over 2$

Thus, the solution of the given equation is $space space space space space space space space space space space space straight x equals 3 over 2$ .