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# How to Solve Equations with Fractions

An equation involving fractions is generally termed as rational equations, in which the unknown variable is to be calculated by solving the equation, in such a way as to find that value of the variable which satisfies the given equation.
The method to solve such equations is known as a cross multiplication method.

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## How to Solve Equations with Fractions

Cross Multiplication Method:
We know, that a linear equation is of the form of ax + b = c, where a, b and c are real numbers and x is a variable, also a not equal to 0, but some equations which are not linear such as $\frac{ax+b}{cx+d}=k$, where a, b, c, d and k are real numbers and cx + d is not equal to zero, can be reduced to the above linear equations.

The method adopted here, for this purpose is called cross multiplication.

In the method of Cross multiplication, the numerator on the left hand side (LHS) of the equation is multiplied by its denominator on the right hand side (RHS) and the denominator on the LHS is multiplied with the denominator on the RHS.

If we have an equation of the form of $\frac{a_{x}}{b_{x}}= \frac{c}{d}$, (here, $\frac{a_{x}}{b_{x}}$ and $\frac{c}{d}$ are in the form of fractions) where a, b, c, d are real numbers bx and d not equal to zero, then the Steps to solve this equation with fractions are:
Step 1:  Always simplify the equation to get a form of equation like $\frac{a_{x}}{b_{x}}= \frac{c}{d}$.
Step 2: Apply the method of cross multiplication, by multiplying ax with d and bx with c.
Step 3: Solve ax* d = bx* c to get a linear equation of the form of ax + b = c.
Step 4: Solve for x, by taking all variables to LHS and the constants to the RHS, and substitute the value or values of x accordingly into the given equation to check whether LHS is equal to RHS.
Step 5: Check whether LHS = RHS, if yes, the required answer is correct, and if no, the answer is not correct, which needs to be reevaluated from the beginning of the step1.

Solved Examples:
Example 1: Solve $\frac{5x-8}{3x}=2$.
Solution: $\frac{5x-8}{3x}=2$ can be rewritten as $\frac{5x-8}{3x}=\frac{2}{1}$.
Applying cross multiplication, we get (5x – 8 ) * 1 = 2 * 3x, this implies 5x – 8 = 6x.
Taking like terms, on LHS and constants on RHS, we get 5x – 6x = 8, this implies –x = 8, and Hence, x = -8 is the required answer.
We can substitute this value, x = -8 in the above given equation to check LHS = RHS.

Example 2: Solve $\frac{3}{4}y+\frac{7}{5}y = 15$
Solution: $\frac{3}{4}y+\frac{7}{5}y = 15$
$\frac{15y+28y}{20}=15$
$\frac{43y}{20}=15$
On cross multiplying we will get, 43y = 300
So y = $\frac{300}{43}$

Example 3: Solve $\frac{2x+3}{6}= 15$
Solution: On cross multiplying we will get, 2x + 3 = 90
2x = 87
x = $\frac{87}{2}$

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