Question

Please solve it I have exam

Answer 1

Answer:

Question :-

$\mapsto \bf{Solve :\: \dfrac{3x - 2}{2x - 3} =\: \dfrac{3x - 8}{x + 4}}$

To Find :-

• What is the value of x.

Solution :-

$\dashrightarrow \sf\bold{\purple{\dfrac{3x - 2}{2x - 3} =\: \dfrac{3x - 8}{x + 4}}}$

$\implies \sf \dfrac{3x - 2}{2x - 3} =\: \dfrac{3x - 8}{x + 4}$

$\pink{\bigstar}\: \: \bf{By\: doing\: cross\: multiplication\: we\: get\: :-}$

$\implies \sf (3x - 2)(x + 4) =\: (2x - 3)(3x - 8)$

$\implies \sf 3x^2 + 12x - 2x - 8 =\: 6x^2 - 16x - 9x + 24$

$\implies \sf 6x^2 - 3x^2 - 16x - 9x - 12x + 2x + 24 + 8 =\: 0$

$\implies \sf 3x^2 - 25x - 10x + 32 =\: 0$

$\implies \sf 3x^2 - 35x + 32 =\: 0$

$\implies \sf 3x^2 - (32 + 3)x + 32 =\: 0$

$\pink{\bigstar}\: \: \bf{By\: splitting\: the\: middle\: term\: we\: get\: :-}$

$\implies \sf 3x^2 - 32x - 3x + 32 =\: 0$

$\implies \sf x(3x - 32) - 1(3x - 32) =\: 0$

$\implies \sf (x - 1)(3x - 32) =\: 0$

$\longrightarrow \sf x - 1 =\: 0$

$\longrightarrow \sf\bold{\red{x =\: 1}}$

Either,

$\longrightarrow \sf 3x - 32 =\: 0$

$\longrightarrow \sf 3x =\: 32$

$\longrightarrow \sf\bold{\red{x =\: \dfrac{32}{3}}}$

${\small{\bold{\underline{\therefore\: The\: value\: of\: x\: is\: 1\: or\: \dfrac{32}{3}\: .}}}}$

Answer 2

Step-by-step explanation:

Given:-

$\rm \: \frac{3x - 2}{2x - 3 } = \frac{3x - 8}{x + 4}$

To find out:-

Value of x .

Solution:-

We have

$\rm \: \frac{3x - 2}{2x - 3 } = \frac{3x - 8}{x + 4}$

Variable x cannot be equal to any of the value -4, 3/2 since dividing by zero is not defined. Multiply both sides on the equation by (2x - 3)(x + 4), the last common multiple of 2x - 3 , x + 4.

↦$\rm \: (x + 3)(3x - 2) = (2x - 3)(3x - 8)$

Use the distribution property to multiply x + 4 by 3x -2 and combine like terms.

↦$\rm {3x}^{2} + 10x - 8 = (2x - 3)(3x - 8)$

Use the distribution property to multiply 2x -3 by 3x -8 and combine like terms.

↦$\rm {3x}^{2} + 10x - 8 = {6x}^{2} - 25x + 24$

Subtract 6x from both sides

↦$\rm {3x}^{2} + 10x - 8 - {6x}^{2} = - 25x + 24$

Add 24 to both sides.

Combine 10x and 25x to get 35x

↦$\rm { - 3x}^{2} + 35x - 8 = 24$

Subtract 24 from both sides

↦$\rm { - 3x}^{2} + 35x - 8 - 24 = 0$

Subtract 24 from -8 to get -32

↦$\rm { - 3x}^{2} + 35x - 32 = 0$

This equation is in standard form of ax² + bx + c = 0.

Substituting -3 from a , 35 from b and -32 from c in quadratic formula,$\boxed{ \rm \frac{ - b± \sqrt{ {b - 4ac}^{2} } }{2a}}$

$\rm x = \frac{ - 35± \sqrt{35 - 4( - 3)( - 32)} }{2( - 3)}$

Square 35.

$\rm x = \frac{ - 35± \sqrt{1225 - 4( - 3)( - 32)} }{2( - 3)}$

Multiply -4 times -3.

$\rm x = \frac{ - 35± \sqrt{1225 + 12( - 32)} }{2( - 3)}$

Multiply 12 times -32.

$\rm x = \frac{ - 35± \sqrt{1225 - 384} }{2( - 3)}$

Add 1225 to -384.

$\rm x = \frac{ - 35± \sqrt{841} }{2( - 3)}$

Take the square root of 841.

$\rm x = \frac{ - 35± 29}{2( - 3)}$

Multiply 2 times -3 in numerator.

$\rm x = \frac{ - 35±29 }{ - 6}$

Now solve the equation $\rm x = \frac{ - 35±29 }{ - 6}$ when ± is plus.

Add -35 to 29.

$\rm \: x = \frac{ - 6}{6}$

Divide -6 by -6

$\rm x = 1$

Now, solve the equation $\rm x = \frac{ - 35±29 }{ - 6}$ when ± is minus.

Subtract 29 from -35.

$\rm x = \frac{ - 61}{ - 6}$

Reduce the fraction-64/-6 to the lower term by extracting and cancaling out 2.

$\rm x = \frac{32}{3}$

Answer:-

$\rm x = 1,x = \frac{32}{3} = 10 \frac{2}{3}$