Question

Solve and check:  3 x2 + 5x + 6 + x − 1 x + 2 = 7 x + 3 x = is a solution. x = is an extraneous solution.

Answer 1

Answer:

• x = -2 is an extraneous solution of $\frac{3}{x^2+5x+6}+\frac{x-1}{x+2}=\frac{7}{x+3}$

• $x = 7$ is the solution of $\frac{3}{x^2+5x+6}+\frac{x-1}{x+2}=\frac{7}{x+3}$.

Step-by-step explanation:

As the given expression

$\frac{3}{x^2+5x+6}+\frac{x-1}{x+2}=\frac{7}{x+3}$

Lets solve this expression

$\frac{3}{(x+3)(x+2)}+\frac{x-1}{x+2}=\frac{7}{x+3}$

Multiply all terms by (x+2)(x+3) and cancel:

$3+(x-1)(x+3)=7(x+2)$

$x^{2} +2x=7x+14$

$x^{2} +2x-(7x+14)=7x+14-(7x+14)$

$x^{2} -5x-14=0$

$(x+2)(x-7)=0$

$x+2=0$  or  $x-7=0$

$x = -2$  or  $x = 7$

x = -2 is an extraneous solution while  $x = 7$ is the right solution.

Verification:

Check answers. (Plug them in to make sure they work.)

Putting $x = -2$

$\frac{3}{x^2+5x+6}+\frac{x-1}{x+2}=\frac{7}{x+3}$

$\frac{3}{(-2)^2+5(-2)+6}+\frac{(-2)-1}{(-2)+2}=\frac{7}{(-2)+3}$

$\frac{3}{4-10+6}+\frac{-3}{0}=\frac{7}{1}$

$\frac{3}{0}+\frac{-3}{0}=\frac{7}{1}$

∞ + ∞ = 7  

L.H.S ≠ R.H.S

x = -2 is an extraneous solution.

Putting $x = 7$

$\frac{3}{x^2+5x+6}+\frac{x-1}{x+2}=\frac{7}{x+3}$

$\frac{3}{(7)^2+5(7)+6}+\frac{(7)-1}{(7)+2}=\frac{7}{(7)+3}$

$\frac{3}{49+35+6}+\frac{6}{9}=\frac{7}{10}$

$\frac{3}{90}+\frac{6}{9}=\frac{7}{10}$

$\frac{3+60}{90}=\frac{7}{10}$

$\frac{63}{90}=\frac{7}{10}$

$\frac{7}{10}=\frac{7}{10}$

L.H.S = R.H.S

So, $x = 7$ is the solution of $\frac{3}{x^2+5x+6}+\frac{x-1}{x+2}=\frac{7}{x+3}$.

Keywords: solution, extraneous solution, equation

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Answer 2

Answer:

1. x=7

2. x=-2

Step-by-step explanation: