gabrielle0409
Answered

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solve for rational algebraic equation.

Solve For Rational Algebraic Equation class=

Sagot :

Step-by-step explanation:

Taking the provided equation ,

[tex]\implies \dfrac{3x+4}{5}-\dfrac{2}{x+3} =\dfrac{8}{5} [/tex]

1) Here denominator of two fractions are 5 and x +3 . So their LCM will be 5(x+3)

[tex]\implies \dfrac{(x+3)(3x+4)-(2)(5)}{5(x+3)}=\dfrac{8}{5} [/tex]

2) Transposing 5(x+3) to Right Hand Side . And 5 to Left Hand Side .

[tex]\implies 5(3x^2+4x+9x+12 -10) = 40(x+3)[/tex]

3) Multiplying the expressions.

[tex] \implies 5(3x^2+13x+2) = 40x + 120 [/tex]

4) Opening the brackets .

[tex]\implies 15x^2+ 65x + 10 = 40x + 120 [/tex]

5) Transposing all terms to Left Hand Side .

[tex]\implies 15x^2 + 65x - 40x + 10 - 120 = 0 \\\\\implies 15x^2 +25x - 110 = 0[/tex]

6) Solving the quadratic equation .

[tex]\implies 5(3x^2+5x -22) = 0 \\\\\implies 3x^2+13x-22 = 0 \\\\ \implies x = \dfrac{-b\pm \sqrt{b^2-4ac}}{4ac} \\\\\implies x = \dfrac{-5\pm \sqrt{5^2-4(-22)(3)}}{2(3)} \\\\\implies x = \dfrac{-5\pm \sqrt{289}}{6}\\\\\implies x =\dfrac{-5\pm 17}{6} \\\\\implies x = \dfrac{17-5}{6},\dfrac{-17-5}{6}\\\\\implies x = \dfrac{12}{6},\dfrac{-22}{6} \\\\\underline{\boxed{\red{\bf\implies x = 2 , \dfrac{-11}{3}}}}[/tex]

elcharly64

Answer:

[tex]x=2\\x=-\frac{11}{3}[/tex]

Step-by-step explanation:

Solve the rational equation:

[tex]\displaystyle \frac{3x+4}{5}-\frac{2}{x+3}=\frac{8}{5}[/tex]

To eliminate denominators, multiply by 5(x+3) (x cannot have a value of 3):

[tex]\displaystyle 5(x+3)\frac{3x+4}{5}-5(x+3)\frac{2}{x+3}=5(x+3)\frac{8}{5}[/tex]

Operate and simplify:

[tex]\displaystyle (x+3)(3x+4)-5(2)=(x+3)(8)[/tex]

[tex]\displaystyle 3x^2+4x+9x+12-10=8x+24[/tex]

Rearranging:

[tex]\displaystyle 3x^2+4x+9x+12-10-8x-24=0[/tex]

Simplifying:

[tex]\displaystyle 3x^2+5x-22=0[/tex]

Rewrite:

[tex]\displaystyle 3x^2-6x+11x-22=0[/tex]

Factoring:

[tex]\displaystyle 3x(x-2)+11(x-2)=0[/tex]

[tex]\displaystyle (x-2)(3x+11)=0[/tex]

Solving:

[tex]x=2\\x=-\frac{11}{3}[/tex]

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