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1/(x-5)+3/(x+2)=4

Sagot :

fieryanswererft

Answer:

[tex]\boxed{ \sf x=\dfrac{4\pm\sqrt{43}}{2}}[/tex]

Explanation:

[tex]\rightarrow \sf \dfrac{1}{x-5}+\dfrac{3}{x+2}=4[/tex]

make the denominators same

[tex]\rightarrow \sf \dfrac{1(x+2)}{(x-5)(x+2)}+\dfrac{3(x-5)}{(x+2)(x-5)}=4[/tex]

join the fractions together

[tex]\rightarrow \sf \dfrac{x+2+3x-15}{(x-5)(x+2)}=4[/tex]

cross multiply

[tex]\rightarrow \sf x+2+3x-15=4(x-5)(x+2)[/tex]

simplify

[tex]\rightarrow \sf 4x-13=4x^2-12x-40[/tex]

group the variables

[tex]\rightarrow \sf 4x^2-16x-27 = 0[/tex]

use quadratic formula

[tex]\rightarrow \sf x = \dfrac{-\left(-16\right)\pm \sqrt{\left(-16\right)^2-4\cdot \:4\left(-27\right)}}{2\cdot \:4}[/tex]

simplify the following

[tex]\rightarrow \sf x=\dfrac{4\pm\sqrt{43}}{2}[/tex]

semsee45

Answer:

[tex]x=\dfrac{4 \pm \sqrt{43}}{2}[/tex]

Step-by-step explanation:

Given equation:

[tex]\dfrac{1}{(x-5)}+\dfrac{3}{(x+2)}=4[/tex]

Make the denominators of the algebraic fractions the same, then combine them into one fraction:

[tex]\begin{aligned}\implies \dfrac{1}{(x-5)} \cdot \dfrac{(x+2)}{(x+2)}+\dfrac{3}{(x+2)}\cdot \dfrac{(x-5)}{(x-5)} & =4\\\\\implies \dfrac{x+2}{(x-5)(x+2)}+\dfrac{3(x-5)}{(x-5)(x+2)} & = 4\\\\ \implies \dfrac{x+2+3(x-5)}{(x-5)(x+2)} & = 4\\\\ \implies \dfrac{4x-13}{(x-5)(x+2)} & = 4 \end{aligned}[/tex]

Multiply both sides of the equation by [tex](x-5)(x+2)[/tex]:

[tex]\begin{aligned}\implies \dfrac{(4x-13)}{(x-5)(x+2)}\cdot (x-5)(x+2) & = 4(x-5)(x+2)\\\\\dfrac{(4x-13)(x-5)(x+2)}{(x-5)(x+2)} & = 4(x-5)(x+2)\\\\4x-13 & = 4(x-5)(x+2)\\\\4x-13 & =4x^2-12x-40\\\\4x^2-16x-27 & = 0\end{aligned}[/tex]

Solve using the Quadratic Formula:

[tex]x=\dfrac{-b \pm \sqrt{b^2-4ac} }{2a}\quad\textsf{when}\:ax^2+bx+c=0[/tex]

[tex]\implies a=4\\\implies b=-16\\\implies c=-27[/tex]

Therefore:

[tex]\implies x=\dfrac{-(-16) \pm \sqrt{(-16)^2-4(4)(-27)} }{2(4)}[/tex]

[tex]\implies x=\dfrac{16 \pm \sqrt{688}}{8}[/tex]

[tex]\implies x=\dfrac{16 \pm 4 \sqrt{43}}{8}[/tex]

[tex]\implies x=\dfrac{4 \pm \sqrt{43}}{2}[/tex]

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