Solve 2 x + 3 x + 3 = 3 2 and also verify the solution.

Given, 2 x + 3 x + 3 = 3 2 . By cross-multiplication, we get 2 2 x + 3 = 3 x + 3 ⇒ 4 x + 6 = 3 x + 9 ⇒ 4 x - 3 x = 9 - 6 ⇒ x = 3 Verification: For x = 3 , LHS = 2 x + 3 x + 3 = 2 × 3 + 3 3 + 3 = 9 6 = 3 2 and RHS = 3 2 ∴ LHS = RHS Hence, x = 3 is the solution to the given equation.

Solve 2 x + 3 x + 3 = 3 2 and also verify the solution.

Given, 2 x + 3 x + 3 = 3 2 . By cross-multiplication, we get 2 2 x + 3 = 3 x + 3 ⇒ 4 x + 6 = 3 x + 9 ⇒ 4 x - 3 x = 9 - 6 ⇒ x = 3 Verification: For x = 3 , LHS = 2 x + 3 x + 3 = 2 × 3 + 3 3 + 3 = 9 6 = 3 2 and RHS = 3 2 ∴ LHS = RHS Hence, x = 3 is the solution to the given equation.

Solve 1 + x 1 - x = 2 3 and also verify the solution.

Given, 1 + x 1 - x = 2 3 . By cross-multiplication, we get 3 1 + x = 2 1 - x ⇒ 3 + 3 x = 2 - 2 x Shifting the constant term to RHS and variable to LHS , we get ⇒ 3 x + 2 x = 2 - 3 ⇒ 5 x = - 1 ⇒ x = - 1 5 Verification: For x = - 1 5 , LHS = 1 + x 1 - x = 1 + - 1 5 1 - - 1 5 ⇒ 1 - 1 5 1 + 1 5 = 5 - 1 5 5 + 1 5 ⇒ 4 5 6 5 = 4 6 = 2 3 and RHS = 2 3 ∴ LHS = RHS Hence, x = - 1 5 is the solution to the given equation.

Solve x + 4 x - 4 = 3 2 and also verify the solution.

Given, x + 4 x - 4 = 3 2 . By cross-multiplication, we get 2 x + 4 = 3 x - 4 ⇒ 2 x + 8 = 3 x - 12 Shifting the constant term to RHS and variable to LHS , we get ⇒ 2 x - 3 x = - 12 - 8 ⇒ - x = - 20 ⇒ x = 20 Verification: For x = 20 , LHS = x + 4 x - 4 = 20 + 4 20 - 4 ⇒ 24 16 = 3 2 and RHS = 3 2 ∴ LHS = RHS Hence, x = 20 is the solution to the given equation.

Solve - x 1 - x = 7 5 and also verify the solution.

Given, - x 1 - x = 7 5 . By cross-multiplication, we get 5 - x = 7 1 - x ⇒ - 5 x = 7 - 7 x Shifting the variable term to the LHS , we get ⇒ - 5 x + 7 x = 7 ⇒ 2 x = 7 ⇒ x = 7 2 Verification: For x = 7 2 , LHS = - x 1 - x = - 7 2 1 - 7 2 = - 7 2 2 - 7 2 = - 7 2 - 5 2 = - 7 - 5 = 7 5 and RHS = 7 5 ∴ LHS = RHS Hence, x = 7 2 is the solution to the given equation.

Solve 9 x - 2 4 - 5 + x 3 = 15 x - 158 12 and also verify the solution.

Given, 9 x - 2 4 - 5 + x 3 = 15 x - 158 12 . By taking LCM of denominators of LHS terms and simplify, we get 3 9 x - 2 - 4 5 + x 12 = 15 x - 158 12 ∵ LCM 4 , 3 = 12 ⇒ 27 x - 6 - 20 - 4 x 12 = 15 x - 158 12 ⇒ 23 x - 26 12 = 15 x - 158 12 Denominator of both sides terms are same, so numerators of terms are equal ⇒ 23 x - 26 = 15 x - 158 Shifting the variable terms to LHS and constant term to the RHS , we get ⇒ 23 x - 15 x = - 158 + 26 ⇒ 8 x = - 132 ⇒ x = - 132 8 = - 33 2 ⇒ x = - 16 1 2 Verification: For x = - 16 1 2 , LHS = 9 x - 2 4 - 5 + x 3 = 9 - 16 1 2 - 2 4 - 5 + - 16 1 2 3 = 9 × - 33 2 - 2 4 - 5 + - 33 2 3 = = 9 × - 33 2 - 2 1 4 - 5 1 + - 33 2 3 = - 297 - 4 2 4 - 10 - 33 2 3 = - 301 8 - - 23 6 = - 903 + 92 24 = - 811 24 and RHS = 15 x - 158 12 = 15 - 16 1 2 - 158 12 = 15 - 33 2 - 158 12 = - 495 2 - 158 1 12 = - 495 - 316 2 12 = - 811 24 ∴ LHS = RHS Hence, x = - 16 1 2 is the solution to the given equation.

Solve 2 x + 2 . 5 - 1 5 x - 4 . 2 = 2 + x and also verify the solution.

Given, 2 x + 2 . 5 - 1 5 x - 4 . 2 = 2 + x . Simplify the terms, we get 2 x + 5 - x - 4 . 2 5 = 2 + x ⇒ 2 x + 5 1 - x - 4 . 2 5 = 2 + x By taking LCM and simplify, ⇒ 5 2 x + 5 - x - 4 . 2 5 = 2 + x ∵ LCM of 1 , 5 = 5 ⇒ 10 x + 25 - x + 4 . 2 5 = 2 + x ⇒ 9 x + 29 . 2 5 = 2 + x 1 Cross-multiplying the terms, we get ⇒ 9 x + 29 . 2 = 5 2 + x ⇒ 9 x + 29 . 2 = 10 + 5 x Shifting the constant terms to RHS and variable to LHS , we get ⇒ 9 x - 5 x = - 29 . 2 + 10 ⇒ 4 x = - 19 . 2 ⇒ x = - 19 . 2 4 ⇒ x = - 4 . 8 Verification: For x = - 4 . 8 , LHS = 2 x + 2 . 5 - 1 5 x - 4 . 2 = 2 - 4 . 8 + 2 . 5 - 1 5 - 4 . 8 - 4 . 2 = - 9 . 6 + 5 - - 4 . 8 - 4 . 2 5 = - 4 . 6 + 1 . 8 = - 2 . 8 and RHS = 2 + x = 2 + - 4 . 8 = 2 - 4 . 8 = - 2 . 8 ∴ LHS = RHS Hence, x = - 4 . 8 is the solution to the given equation.