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Please could you explain this step by step I’m really stuck.

Please Could You Explain This Step By Step Im Really Stuck class=

Sagot :

Chetankachanaidn

Answer:

x = -2.5
x = 2

Step-by-step explanation:

Hello!

We can remove the denominators by multiplying everything by it.

Solve:

  • [tex]\frac{5}{x + 3} + \frac4{x + 2} = 2[/tex]
  • [tex]5 + \frac{4(x + 3)}{x + 2} = 2(x + 3)[/tex]
  • [tex]5 + \frac{4x + 12}{x + 2} = 2x + 6[/tex]
  • [tex]5(x + 2) +4x + 12 = (2x + 6)(x + 2)[/tex]
  • [tex]5x + 10 + 4x + 12 = 2x^2 + 6x + 4x + 12[/tex]
  • [tex]9x + 22 = 2x^2 + 10x + 12[/tex]

Let's make one side equal to 0.

  • [tex]0 = 2x^2 + x - 10[/tex]

Solve by factoring, and then the zero product property.

  • [tex]0 = 2x^2 + x - 10\\[/tex]

Multiply 2 and -10. You get -20. Think of two number that multiply to -20, and add to 1. If we think about the standard form of a quadratic, [tex]ax^2 + bx + c[/tex], think two numbers that equal "ac" and add up to "b".

  • [tex]0 = 2x^2 - 4x + 5x - 10[/tex]
  • [tex]0 = 2x(x - 2) + 5(x - 2)[/tex]
  • [tex]0 = (2x + 5)(x - 2)[/tex]

Using the zero product property, set each factor to 0 and solve for x.

  1. 2x + 5 = 0
    2x = -5
    x = -5/2 = -2.5
  2. x - 2 = 0
    x = 2

The two answers are x = -2.5, and x = 2.

VectorFundament120idn

Answer:

x = -5/2, 2

Step-by-step explanation:

Given:

[tex]\displaystyle \large{\dfrac{5}{x+3}+\dfrac{4}{x+2} = 2}[/tex]

With restriction that x-values cannot be -3 and -2 else it will turn the expression as in undefined.

First, multiply both sides by (x+3)(x+2) to get rid of the denominator.

[tex]\displaystyle \large{\dfrac{5}{x+3}\cdot (x+2)(x+3)+\dfrac{4}{x+2} \cdot (x+2)(x+3) = 2 \cdot (x+2)(x+3)}\\\\\displaystyle \large{5(x+2)+4(x+3) = 2(x+2)(x+3)}[/tex]

Simplify/Expand in:

[tex]\displaystyle \large{5x+10+4x+12 = 2(x^2+5x+6)}\\\\\displaystyle \large{9x+22=2x^2+10x+12}[/tex]

Arrange the terms or expression in quadratic equation:

[tex]\displaystyle \large{0=2x^2+10x+12-9x-22}\\\\\displaystyle \large{2x^2+x-10=0}[/tex]

Factor the expression:

[tex]\displaystyle \large{(2x+5)(x-2)=0}[/tex]

Solve like linear equation as we get:

[tex]\displaystyle \large{x=-\dfrac{5}{2}, 2}[/tex]

Since both x-values are not exact -2 or -3 - therefore, these values are valid. Hence, x = -5/2, 2

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