EQUATIONS WITH FRACTIONS

To solve an equation with fraction means to find the numerical value of the unknown such that when the result is substituted into the equation, the equation becomes valid or true.

The next 5 examples illustrate this fact.

Example 1
Solve for $x$ in $\frac{4}{x+3}=\frac{3}{x+2}$

Solution:
First, we cross multiply
$4(x+2)=3(x+3)$
$4x+8=3x+9$

Collecting like terms
$4x-3x=9-8$
$x=1$

Now, let's check if the answer is correct
$\frac{4}{1+3}=\frac{3}{1+2}$
$1=1$... Which is true.

Example 2
Solve $\frac{1}{\frac{1}{2}-1}=\frac{2}{\frac{1}{2}+x}$

Solution:
First, we take the L.C.D of both sides
$\frac{1}{\frac{1-2}{2}}=\frac{2}{\frac{1+2x}{2}}$
$\frac{1}{\frac{-1}{2}}=\frac{2}{\frac{1+2x}{2}}$

Now, we cross multiply
$1\times\frac{1+2x}{2}=2\times\frac{-1}{2}$
$\frac{1+2x}{2}=-1$

Once again, we cross multiply
$-1\times2=1+2x$
$-2=1+2x$
$-2-1=2x$
$-3=2x$

Divide both sides by 2
$\frac{-3}{2}=\frac{2x}{2}$
$x=\frac{-3}{2}$

Example 3
Determine the value of $x$ in $\frac{2x+1}{6}=\frac{3x+1}{4}$

Solution:
First, we cross multiply
$4(2x+1)=6(3x+1)$
$8x+4=18x+6$
$4-6=18x-8x$
$-2=10x$
$\frac{-2}{10}=\frac{10x}{10}$
$x=\frac{-1}{5}$

Example 4
Determine the value of $y$ in
$\frac{4y-1}{y+4}-2=\frac{2y-1}{y+2}$

Solution:
First, let take the L.C.M of the right-hand side
$\frac{4y-1-2(y+4)}{y+4}=\frac{2y-1}{y+2}$
$\frac{4y-1-2y+8}{y+4}=\frac{2y-1}{y+2}$
$\frac{2y-9}{y+4}=\frac{2y-1}{y+2}$

Now, let cross multiply
$(2y-9)(y+2)={2y-1)(y+4)$

Expanding the brackets
$2y(y+2)-9(y+2)=2y(y+4)-1(y+4)$
$2y^2+4y-9y-18=2y^2+8y-y-4$
$2y^2-5y-18=2y^2+7y-4$

Collecting like terms
$2y^2-2y^2-5y-7y=-4+18=$
$-12y=14$
$\frac{-12y}{-12}=\frac{14}{-12}$
$y=\frac{-7}{6}$

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Example 5
Solve $\frac{20}{5x+15}=\frac{15}{5x+10}$

Solution:
$20(5x+10)=15(5x+15)$
$100x+200=75x+225$
$100x-75x=225-200$
$25x=25$
$\frac{25x}{25}=\frac{25}{25}$
$x=1$
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