Question

i need this 2 answers​

Answer 1

~Anzwer :-

• Equation 1)...

$\frac{3x}{2} + 1 = \frac{7}{4} x + 3$

$2\times 3x+4=7x+12$

$6x+4=7x+12$

$6x+4-7x=12$

$-x+4=12$

$-x=12-4$

$-x=8$

$x = - 8$

∴ Hence, the first equation, value of $ x $ will be $ \bf{-8}$.

• Equation 2)...

$\frac{1}{2} (x + 1) + \frac{1}{3} (x - 1) = \frac{5}{12} (x - 2)$

$\frac{1}{2}x+\frac{1}{2}+\frac{1}{3}\left(x-1\right)=\frac{5}{12}\left(x-2\right)$

$\frac{1}{2}x+\frac{1}{2}+\frac{1}{3}x+\frac{1}{3}\left(-1\right)=\frac{5}{12}\left(x-2\right)$

$\frac{1}{2}x+\frac{1}{2}+\frac{1}{3}x-\frac{1}{3}=\frac{5}{12}\left(x-2\right)$

$\frac{5}{6}x+\frac{1}{2}-\frac{1}{3}=\frac{5}{12}\left(x-2\right)$

$\frac{5}{6}x+\frac{3}{6}-\frac{2}{6}=\frac{5}{12}\left(x-2\right)$

$\frac{5}{6}x+\frac{3-2}{6}=\frac{5}{12}\left(x-2\right)$

$\frac{5}{6}x+\frac{1}{6}=\frac{5}{12}\left(x-2\right)$

$\frac{5}{6}x+\frac{1}{6}=\frac{5}{12}x+\frac{5}{12}\left(-2\right)$

$\frac{5}{6}x+\frac{1}{6}=\frac{5}{12}x+\frac{-10}{12}$

$\frac{5}{6}x+\frac{1}{6}=\frac{5}{12}x-\frac{5}{6}$

$\frac{5}{6}x+\frac{1}{6}-\frac{5}{12}x=-\frac{5}{6}$

$\frac{5}{12}x+\frac{1}{6}=-\frac{5}{6}$

$\frac{5}{12}x=-\frac{5}{6}-\frac{1}{6}$

$\frac{5}{12}x=\frac{-5-1}{6}$

$\frac{5}{12}x=\frac{-6}{6}$

$\frac{5}{12}x=-1$

$x=-\frac{12}{5}$

∴ Hence, in the second equation, value of $ x $ will be $ \bf{-\frac{12}{5}}$.

Answer 2

Required answers :-

ㅤㅤㅤㅤ1) x = -8 ㅤ

ㅤㅤㅤㅤ2) x= -12/5ㅤㅤ

Given to solve for x :- ㅤㅤㅤ

$ㅤㅤㅤ \dfrac{3x}{2} + 1 = \dfrac{7}{4} x + 3$

$\dfrac{1}{2} (x + 1) + \dfrac{1}{3} (x - 1) = \dfrac{5}{12} (x - 2)$

SOLUTIONS :-

Question- 1 :-

$ㅤㅤㅤ \dfrac{3x}{2} + 1 = \dfrac{7}{4} x + 3$

‎ㅤㅤTake L.C.M to the denominators in L.H.S and R.H.S

$\dfrac{3x + 2(1)}{2} = \dfrac{7x + 4(3)}{4}$

$\dfrac{3x + 2}{2} = \dfrac{7x + 12}{4}$

$3x + 2 = \dfrac{7x + 12}{2}$

ㅤDo the cross multiplication

$(3x + 2)2 = 7x + 12$

$6x + 4 = 7x + 12$

Transpose R.H.S terms to L.H.S

$6x + 4 - 7x - 12 = 0$

$- x - 8 = 0$

$- x = 8$

$x = \: - 8$

ㅤ So, the value of x is -8

ㅤㅤVerification :-

Substitute value of x it should be L.H.S = R.H.S

$ㅤㅤㅤ \dfrac{3x}{2} + 1 = \dfrac{7}{4} x + 3$

$\dfrac{3( - 8)}{2} + 1 = \dfrac{7( - 8)}{4} + 3$

$\dfrac{ - 24}{2} + 1 = \dfrac{ - 56}{4} + 3$

$\dfrac{ - 24 + 2}{2} = \dfrac{ - 56 + 12}{4}$

$\dfrac{ - 22}{2} = \dfrac{ - 44}{4}$

$- 11 = \: - 11$

ㅤㅤHence verified !

_____________________________

Question- 2:-

$\dfrac{1}{2} (x + 1) + \dfrac{1}{3} (x - 1) = \dfrac{5}{12} (x - 2)$

$\dfrac{x + 1}{2} + \dfrac{x - 1}{3} = \dfrac{5(x - 2)}{12}$

ㅤㅤㅤTake L.C.M to the denominators in L.H.S and R.H.S

$\dfrac{( x + 1)3 + (x - 1)2}{6} = \dfrac{5x - 10}{12}$

$\dfrac{3x + 3 + 2x - 2}{6} = \dfrac{5x - 10}{12}$

$\dfrac{5x + 1}{6} = \dfrac{5x - 10}{12}$

$5x + 1 = \dfrac{5x - 10}{2}$ㅤ

Do the cross multiplication

$(5x + 1)2 = 5x - 10$

$10x + 2 = 5x - 10$

ㅤ Transpose R.H.S terms to L.H.S

$10x - 5x + 2 + 10 = 0$

$5x + 12 = 0$

$5x = - 12$

$x = \dfrac{ - 12}{5}$

ㅤㅤSo, the value of x is -12/5

ㅤㅤVerification:-

Substitute value of x it should be L.H.S = R.H.S

$\dfrac{1}{2} (x + 1) + \dfrac{1}{3} (x - 1) = \dfrac{5}{12} (x - 2)$

$\dfrac{1}{2} \bigg(\dfrac{ - 12}{5} + 1 \bigg) + \dfrac{1}{3} \bigg( \dfrac{ - 12}{5} - 1 \bigg) = \dfrac{5}{12} \bigg( \dfrac{ - 12}{5} - 2 \bigg)$

$\dfrac{1}{2} \bigg(\dfrac{ - 12 + 5}{5} \bigg) + \dfrac{1}{3} \bigg( \dfrac{ - 12 - 5}{5} \bigg) = \dfrac{5}{12} \bigg( \dfrac{ - 12- 10}{5} \bigg)$

$\dfrac{1}{2} \bigg(\dfrac{ - 7}{5} \bigg) + \dfrac{1}{3} \bigg( \dfrac{ - 17}{5} \bigg) = \dfrac{5}{12} \bigg( \dfrac{ - 22 }{5} \bigg)$

$\dfrac{ - 7}{10} - \dfrac{17}{15} = \dfrac{5( - 22)}{12(5)}$

$\dfrac{ - 7(3) - 17(2)}{30} = \dfrac{ - 110}{60}$

$\dfrac{ - 21 - 34}{30} = \dfrac{ - 110}{60}$

$\dfrac{ - 55}{30} = \dfrac{ - 11}{6}$

$\dfrac{ - 11}{6} = \dfrac{ - 11}{6}$

ㅤㅤㅤHence verified !

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