Question

solve x+3/2-3x+1/4=2(x-2)/3-2 ​

Answer 1

Following are the steps for getting the answer:

Given:

$\frac{x+3}{2-3x}+\frac{1}{4}=\frac{2(x-2)}{3-2}$

To find:

Value of x

Solution:

we have to solve the $\frac{x+3}{2-3x}+\frac{1}{4}=\frac{2(x-2)}{3-2}$

we will do the fraction addition from left-hand side

$\frac{4\times(x+3)+1(2-3x)}{(2-3x)\times4} =\frac{2(x-2)}{1} \\\\\frac{(4x+12)+(2-3x)}{(8-12x)} =\frac{2(x-2)}{1} \\\\\\\frac{(4x+12+2-3x)}{(8-12x)} =\frac{2(x-2)}{1} \\\\\\\frac{(x+14)}{(8-12x)} =\frac{2(x-2)}{1}$

Now, do the cross multiplicatiom

1(x+14)=(8-12x) 2(x-2)

x+14=(8-12x) (2x-4)

x+14=16x-32-24x²+48x

x+14=64x-32-24x²

x+14=-24x²+64x-32

x=-24x²+64x-32-14

x=-24x²+64x-46

now take -2 as a common

x=-2(12x²-32x+23)

the equation is in the form of quadratic equation.

Thus, the answer is x=-2(12x²-32x+23)

Answer 2

Answer:

The value is $x=5$.

Step-by-step explanation:

Given equation is $\frac{x+3}{2}-\frac{3x+1}{4}=\frac{2(x-2)}{3}-2$.

To solve this, we perform L.C.M. on both sides and make two terms into a single term. Then we do the cross product to find the value of $x$.

On the left side, we need to add both the numbers.

To add these numbers, the denominator should be same.

So the L.C.M. of $2,4$ is $4$.

The equation becomes

$\frac{2(x+3)-(3x+1)}{4}=\frac{2(x-2)}{3}-\frac{2}{1}$

On the right side, the L.C.M. of $3,1$ is $3$.

The equation becomes

$\frac{2(x+3)-(3x+1)}{4}=\frac{2(x-2)-3(2)}{3}$

Expand the numerator on the both sides, we get

$\frac{2x+6-3x-1}{4}=\frac{2x-4-6}{3}$

$\frac{-x+5}{4}=\frac{2x-10}{3}$

Now doing the cross product, we get

$3(-x+5)=4(2x-10)\\-3x+15=8x-40\\8x+3x=15+40\\11x=55\\x=5$

The value of $x$ is $5$.