Question

solve
linear equations
RSAgarrwal​

Answer 1

Answer:

$\textbf{Given:–} \frac{x - 5}{2} - \frac{x - 3}{5} = \frac{1}{2}$

$\sf \colorbox{pink} {Solution:} \: \frac{x - 5}{2} - \frac{x - 3}{5} = \frac{1}{2}$

$⇒ \frac{5(x - 5) - 2(x - 3)}{10} = \frac{1}{2}$

$⇒5x - 25 - 2x + 6 = \frac{10}{2}$

$⇒3x - 19 = 5$

$⇒3x = 19 + 5$

$⇒3x = 24$

$⇒x = \frac{24}{3}$

$⇒x = 8$

$\\ \\ \\ \\ \\ \\ \sf \colorbox{gold} {\red(ANSWER ᵇʸ ⁿᵃʷᵃᵇ⁰⁰⁰⁸}$

Answer 2

☯GIVEN :

• $\bf {\dfrac{x - 5}{2} - \dfrac{x - 3}{5} = \dfrac{1}{2} }$

☯TO FIND :

• The value of x.

➲SOLUTION :

$\implies{ \sf {\dfrac{x - 5}{2} - \dfrac{x - 3}{5} = \dfrac{1}{2} }}$

$\implies{\sf {\dfrac{5 \big(x - 5 \big) - 2 \big(x - 3 \big)}{10} = \dfrac{1}{2} }}$

$\implies{\sf {\dfrac{5x - 25 - 2x + 6 }{10} = \dfrac{1}{2} }}$

$\implies{\sf {\dfrac{5x - 2x + 6 - 25 }{10} = \dfrac{1}{2} }}$

$\implies{\sf {\dfrac{3x - 19 }{10} = \dfrac{1}{2} }}$

• By cross multiplication :

$\implies{ \sf{2 \big(3x - 19 \big) = 10 \times 1} }$

$\implies{\sf{6x -38 = 10} }$

$\implies{\sf{6x = 10 + 38} }$

$\implies{ \sf{6x = 48} }$

$\implies{ \sf{x = \cancel{\dfrac{48}{6}}} }$

$\implies{ \huge {\underline{\boxed { \red{\mathfrak{x = 8}}}}}}$

$\huge{\green{\therefore}}$ The value of x is 8.

✞VERIFICATION :

• Substituting the value of x :

$\implies{ \sf {\dfrac{8 - 5}{2} - \dfrac{8 - 3}{5} = \dfrac{1}{2} }}$

$\implies{ \sf {\dfrac{3}{2} - \dfrac{5}{5} = \dfrac{1}{2} }}$

$\implies{ \sf {\dfrac{15-10}{10}= \dfrac{1}{2} }}$

$\implies{ \sf {\cancel{\dfrac{5}{10}}= \dfrac{1}{2} }}$

$\implies{ \sf {\dfrac{1}{2} = \dfrac{1}{2} }}$

$\implies{ \huge {\underline{\boxed { \red{\mathfrak{LHS=RHS}}}}}}$

$\huge{\green{\therefore}}$ Verified.