June 2026

Solve the Following Quadratic Equation by Factorization Question: \[ \frac{3}{x+1}+\frac{4}{x-1}=\frac{29}{4x-1}, \qquad x\ne 1,-1,\frac14 \] Solution Given: \[ \frac{3}{x+1}+\frac{4}{x-1}=\frac{29}{4x-1} \] Multiplying both sides by \((x+1)(x-1)(4x-1)\): \[ 3(x-1)(4x-1)+4(x+1)(4x-1) =29(x+1)(x-1) \] \[ 3(4x^2-5x+1)+4(4x^2+3x-1) =29(x^2-1) \] \[ 12x^2-15x+3+16x^2+12x-4 =29x^2-29 \] \[ 28x^2-3x-1 =29x^2-29 \] \[ x^2+3x-28=0 \] Factorizing: \[ x^2+7x-4x-28=0 \] \[ x(x+7)-4(x+7)=0 \] \[ (x+7)(x-4)=0 \] Therefore, \[ […]

Solve the Following Quadratic Equation by Factorization Question: \[ \frac{3}{x+1}-\frac{1}{2}=\frac{2}{3x-1}, \qquad x\ne -1,\frac{1}{3} \] Solution Given: \[ \frac{3}{x+1}-\frac{1}{2}=\frac{2}{3x-1} \] Multiplying both sides by \(2(x+1)(3x-1)\): \[ 6(3x-1)-(x+1)(3x-1)=4(x+1) \] \[ 18x-6-(3x^2+2x-1)=4x+4 \] \[ -3x^2+16x-5=4x+4 \] \[ -3x^2+12x-9=0 \] \[ 3x^2-12x+9=0 \] \[ x^2-4x+3=0 \] Factorizing: \[ x^2-3x-x+3=0 \] \[ x(x-3)-1(x-3)=0 \] \[ (x-3)(x-1)=0 \] Therefore, \[ x-3=0

Solve the Following Quadratic Equation by Factorization Question: \[ \frac{5+x}{5-x}-\frac{5-x}{5+x}=\frac{15}{4}, \qquad x\ne 5,-5 \] Solution Given: \[ \frac{5+x}{5-x}-\frac{5-x}{5+x}=\frac{15}{4} \] Taking LCM on the left side: \[ \frac{(5+x)^2-(5-x)^2}{(5-x)(5+x)} =\frac{15}{4} \] Using the identity \(a^2-b^2=(a-b)(a+b)\): \[ \frac{\big[(5+x)-(5-x)\big]\big[(5+x)+(5-x)\big]} {25-x^2} =\frac{15}{4} \] \[ \frac{(2x)(10)}{25-x^2} =\frac{15}{4} \] \[ \frac{20x}{25-x^2} =\frac{15}{4} \] Cross-multiplying: \[ 80x=15(25-x^2) \] \[ 80x=375-15x^2 \] \[ 15x^2+80x-375=0

Solve the Following Quadratic Equation by Factorization Question: \[ \frac{x-2}{x-3}+\frac{x-4}{x-5}=\frac{10}{3}, \qquad x\ne 3,5 \] Solution Given: \[ \frac{x-2}{x-3}+\frac{x-4}{x-5}=\frac{10}{3} \] Multiplying both sides by \(3(x-3)(x-5)\): \[ 3(x-2)(x-5)+3(x-4)(x-3) =10(x-3)(x-5) \] \[ 3(x^2-7x+10)+3(x^2-7x+12) =10(x^2-8x+15) \] \[ 6x^2-42x+66 =10x^2-80x+150 \] \[ 4x^2-38x+84=0 \] \[ 2x^2-19x+42=0 \] Factorizing: \[ 2x^2-12x-7x+42=0 \] \[ 2x(x-6)-7(x-6)=0 \] \[ (x-6)(2x-7)=0 \] Therefore, \[ x-6=0

Solve the Following Quadratic Equation by Factorization Question: \[ \frac{x-4}{x-5}+\frac{x-6}{x-7}=\frac{10}{3}, \qquad x\ne 5,7 \] Solution Given: \[ \frac{x-4}{x-5}+\frac{x-6}{x-7}=\frac{10}{3} \] Cross-multiplying by \(3(x-5)(x-7)\): \[ 3(x-4)(x-7)+3(x-6)(x-5)=10(x-5)(x-7) \] \[ 3(x^2-11x+28)+3(x^2-11x+30) =10(x^2-12x+35) \] \[ 6x^2-66x+174 =10x^2-120x+350 \] \[ 4x^2-54x+176=0 \] \[ 2x^2-27x+88=0 \] Factorizing: \[ 2x^2-16x-11x+88=0 \] \[ 2x(x-8)-11(x-8)=0 \] \[ (x-8)(2x-11)=0 \] Therefore, \[ x-8=0 \quad \text{or} \quad

Solve the Following Quadratic Equation by Factorization Question: \[ \frac{4}{x}-3=\frac{5}{2x+3}, \qquad x\ne 0,-\frac{3}{2} \] Solution Given: \[ \frac{4}{x}-3=\frac{5}{2x+3} \] Multiplying both sides by \(x(2x+3)\): \[ 4(2x+3)-3x(2x+3)=5x \] \[ 8x+12-6x^2-9x=5x \] \[ -6x^2-6x+12=0 \] \[ x^2+x-2=0 \] Factorizing: \[ x^2+2x-x-2=0 \] \[ x(x+2)-1(x+2)=0 \] \[ (x+2)(x-1)=0 \] Therefore, \[ x+2=0 \quad \text{or} \quad x-1=0 \] \[

Solve the Following Quadratic Equation by Factorization Question: \[ \frac{x-1}{2x+1}+\frac{2x+1}{x-1}=\frac{5}{2}, \qquad x\ne -\frac12,\;1 \] Solution Given: \[ \frac{x-1}{2x+1}+\frac{2x+1}{x-1}=\frac{5}{2} \] Taking LCM on the left side: \[ \frac{(x-1)^2+(2x+1)^2}{(2x+1)(x-1)} =\frac{5}{2} \] Cross-multiplying: \[ 2\Big[(x-1)^2+(2x+1)^2\Big] =5(2x+1)(x-1) \] \[ 2\Big[(x^2-2x+1)+(4x^2+4x+1)\Big] =5(2x^2-x-1) \] \[ 10x^2+4x+4 =10x^2-5x-5 \] \[ 9x+9=0 \] \[ x+1=0 \] \[ x=-1 \] Since \(x=-1\) does not

Solve the Following Quadratic Equation by Factorization Question: \[ \frac{x+1}{x-1}-\frac{x-1}{x+1}=\frac{5}{6}, \qquad x\ne 1,-1 \] Solution Given: \[ \frac{x+1}{x-1}-\frac{x-1}{x+1}=\frac{5}{6} \] Taking LCM on the left side: \[ \frac{(x+1)^2-(x-1)^2}{(x-1)(x+1)} =\frac{5}{6} \] Using the identity \(a^2-b^2=(a-b)(a+b)\): \[ \frac{\big[(x+1)-(x-1)\big]\big[(x+1)+(x-1)\big]} {x^2-1} =\frac{5}{6} \] \[ \frac{(2)(2x)}{x^2-1} =\frac{5}{6} \] \[ \frac{4x}{x^2-1} =\frac{5}{6} \] Cross-multiplying: \[ 24x=5(x^2-1) \] \[ 24x=5x^2-5 \] \[ 5x^2-24x-5=0

Solve the Following Quadratic Equation by Factorization Question: \[ \frac{1}{x-2}+\frac{2}{x-1}=\frac{6}{x}, \qquad x\ne 0,1,2 \] Solution Given: \[ \frac{1}{x-2}+\frac{2}{x-1}=\frac{6}{x} \] Multiplying both sides by \(x(x-1)(x-2)\): \[ x(x-1)+2x(x-2)=6(x-1)(x-2) \] \[ x^2-x+2x^2-4x=6(x^2-3x+2) \] \[ 3x^2-5x=6x^2-18x+12 \] \[ 3x^2-13x+12=0 \] Factorizing: \[ 3x^2-9x-4x+12=0 \] \[ 3x(x-3)-4(x-3)=0 \] \[ (x-3)(3x-4)=0 \] Therefore, \[ x-3=0 \quad \text{or} \quad 3x-4=0 \] \[

Solve the Following Quadratic Equation by Factorization Question: \[ \frac{x-3}{x+3}-\frac{x+3}{x-3}=\frac{48}{7}, \qquad x\ne 3,-3 \] Solution Given: \[ \frac{x-3}{x+3}-\frac{x+3}{x-3}=\frac{48}{7} \] Taking LCM on the left side: \[ \frac{(x-3)^2-(x+3)^2}{(x+3)(x-3)}=\frac{48}{7} \] Using \(a^2-b^2=(a-b)(a+b)\): \[ \frac{\big[(x-3)-(x+3)\big]\big[(x-3)+(x+3)\big]}{x^2-9} =\frac{48}{7} \] \[ \frac{(-6)(2x)}{x^2-9} =\frac{48}{7} \] \[ \frac{-12x}{x^2-9} =\frac{48}{7} \] Cross-multiplying: \[ -84x=48(x^2-9) \] \[ -84x=48x^2-432 \] \[ 48x^2+84x-432=0 \] Dividing by 12: