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Solve it

[tex] \rm \dashrightarrow \: \frac{7(x - 3)}{4(x + 5)} = \frac{3(x - 5)}{5(x + 4)} \\ [/tex]


Sagot :

msm555

Answer:[tex]\frac{7(x-3)}{4(x+5)}=\frac{3(x-5)}{5(x+4)}[/tex]

Doing criss cross multiplication

7(x-3)*5(x+4)=3(x-5)*4(x+5)

35(x*x+4*x-3*x-3*5)=20(x²-25)

35(x²+4x-3x-15)=20(x²-25)

35(x²+x-15)=20(x²-25)

35/20*(x²+x-15)=x²-25

7/4*(x²+x-15)=x²-25

7x²+7x-7*15=4x²-25*4

7x²-7x-105=4x²-100

7x²-4x²-7x-105+100=0

3x²-7x-5=0

Comparing above equation with ax²+bx+c=0

we get

a=3

b=-7

c=-5

now

By using quadratic equation formula

x=[tex]\frac{ -b±\sqrt{b²-4a}}{2a}[/tex]

Substitute value

x=[tex]\frac{ 7±\sqrt{-7²-4*3}}{2*3}[/tex]

x=[tex]\frac{ 7±\sqrt{61}}{6}[/tex]

taking positive

x=[tex]\frac{ 7+\sqrt{61}}{6}[/tex]

taking negative

x=[tex]\frac{ 7-\sqrt{61}}{6}[/tex]

The solutions to the values of x are [tex]x=\frac{-35 \pm 85.79 }{46} \\[/tex]

Equations and expressions

Given the equation

[tex]\frac{7(x-3)}{4(x+5)} =\frac{3(x-5)}{5(x+4)}[/tex]

Cross multiply

(7x - 21)(5x + 20) = (4x+20)(3x-15)

Expand the function

[tex]35x^2+140x - 105x - 420 = 12x^2 - 60x + 60x - 350\\35x^2 + 35x - 420 = 12x^2 - 300[/tex]

Collect the like terms

[tex]35x^2 - 12x^2 + 35x - 80 = 0\\23x^2 + 35x - 80 = 0[/tex]

Factorizing the result, the value of x

[tex]x=\frac{-35 \pm \sqrt{7360} }{46} \\x=\frac{-35 \pm 85.79 }{46} \\[/tex]

Hence the solution to the values of x are [tex]x=\frac{-35 \pm 87.55 }{46} \\[/tex]

Learn more on quadratic equation here: https://brainly.com/question/1214333