Question

(6y+1)/2+1=(7y-3)/3. solve and check also

Answer 1

like it this question

Answer 2

Answer:

$y=-\frac{15}{4}$

Step-by-step explanation:

The given equation is

$\frac{6y+1}{2}+1=\frac{7y-3}{3}$

Multiply both sides by 6, to simplify the given equation.

$(\frac{6y+1}{2}+1)\times 6=\frac{7y-3}{3}\times 6$

$(\frac{6y+1}{2})6+6=\frac{7y-3}{3}\times 6$

Cancel out common factors.

$(6y+1)3+6=(7y-3)\times 2$

Using distributive property.

$18y+3+6=14y-6$

$18y+9=14y-6$

Isolate variables on left side.

$18y-14y=-9-6$

$4y=-15$

Divide both sides by 4.

$y=-\frac{15}{4}$

Therefore the answer is $y=-\frac{15}{4}$.

Substitute the value of y in the given equation to check whether it is correct or not.

$\frac{6(-\frac{15}{4})+1}{2}+1=\frac{7(-\frac{15}{4})-3}{3}$

$\frac{-\frac{45}{2}+1}{2}+1=\frac{-\frac{105}{4})-3}{3}$

$\frac{-\frac{43}{2}}{2}+1=\frac{-\frac{117}{4}}{3}$

$-\frac{43}{4}+1=-\frac{117}{12}$

$-\frac{39}{4}=-\frac{39}{4}$

$LHS=RHS$

Therefore the answer is correct.