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Fraction cross multiplication Calculator

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1

Solved example of fraction cross multiplication

$\frac{x+\frac{3}{6}}{1-\frac{3x}{6}}=\frac{2-x}{1+2x}$
2

Multiplying the fraction by $-1$

$\frac{x+\frac{1}{2}}{1+\frac{-3x}{6}}=\frac{2-x}{1+2x}$
3

Take $\frac{-3}{6}$ out of the fraction

$\frac{x+\frac{1}{2}}{1-\frac{1}{2}x}=\frac{2-x}{1+2x}$
4

Apply fraction cross-multiplication

$\left(x+\frac{1}{2}\right)\left(1+2x\right)=\left(1-\frac{1}{2}x\right)\left(2-x\right)$
5

Grouping all terms to the left side of the equation

$\left(x+\frac{1}{2}\right)\left(1+2x\right)-\left(1-\frac{1}{2}x\right)\left(2-x\right)=0$
6

Solve the product $-\left(1-\frac{1}{2}x\right)\left(2-x\right)$

$\left(x+\frac{1}{2}\right)\left(1+2x\right)+\left(-1+\frac{1}{2}x\right)\left(2-x\right)=0$
7

Multiplying polynomials $x+\frac{1}{2}$ and $1+2x$

$x+\frac{1}{2}+2x\left(x+\frac{1}{2}\right)+\left(-1+\frac{1}{2}x\right)\left(2-x\right)=0$
8

Solve the product $2x\left(x+\frac{1}{2}\right)$

$x+\frac{1}{2}+x\left(2x+1\right)+\left(-1+\frac{1}{2}x\right)\left(2-x\right)=0$
9

Solve the product $x\left(2x+1\right)$

$x+\frac{1}{2}+2x^2+x+\left(-1+\frac{1}{2}x\right)\left(2-x\right)=0$
10

Multiplying polynomials $-1+\frac{1}{2}x$ and $2-x$

$x+\frac{1}{2}+2x^2+x+2\left(-1+\frac{1}{2}x\right)-x\left(-1+\frac{1}{2}x\right)=0$
11

Solve the product $2\left(-1+\frac{1}{2}x\right)$

$-\frac{3}{2}+x+2x^2+x+x-x\left(-1+\frac{1}{2}x\right)=0$
12

Solve the product $-x\left(-1+\frac{1}{2}x\right)$

$-\frac{3}{2}+x+2x^2+x+x+x\left(1-\frac{1}{2}x\right)=0$
13

Solve the product $x\left(1-\frac{1}{2}x\right)$

$-\frac{3}{2}+x+2x^2+x+x+x-\frac{1}{2}x^2=0$
14

Adding $x$ and $x$

$2x-\frac{3}{2}+2x^2+x+x-\frac{1}{2}x^2=0$
15

Adding $2x^2$ and $-\frac{1}{2}x^2$

$\frac{3}{2}x^2+2x-\frac{3}{2}+x+x=0$
16

Adding $2x$ and $x$

$3x+\frac{3}{2}x^2-\frac{3}{2}+x=0$
17

Adding $3x$ and $x$

$4x+\frac{3}{2}x^2-\frac{3}{2}=0$
18

To find the roots of a polynomial of the form $ax^2+bx+c$ we use the quadratic formula, where in this case $a=\frac{3}{2}$, $b=4$ and $c=-\frac{3}{2}$. Then substitute the values of the coefficients of the equation in the quadratic formula:

  • $\displaystyle x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$

$x=\frac{-4\pm 5}{3}$
19

To obtain the two solutions, divide the equation in two equations, one when $\pm$ is positive ($+$), and another when $\pm$ is negative ($-$)

$x=\frac{-4+5}{3},\:x=\frac{-4-5}{3}$
20

Add the values $-4$ and $5$

$x=\frac{1}{3},\:x=\frac{-4-5}{3}$
21

Add the values $-4$ and $-5$

$x=\frac{1}{3},\:x=\frac{-9}{3}$
22

Solve the equation ($1$)

$x=\frac{1}{3}$
23

Divide $1$ by $3$

$x=\frac{1}{3}$
24

Solve the equation ($2$)

$x=\frac{-9}{3}$
25

Divide $-9$ by $3$

$x=-3$
26

The $2$ solutions of the equation are

$x=\frac{1}{3},\:x=-3$

Answer

$x=\frac{1}{3},\:x=-3$

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