Question

equations are useful to help you solve some real - life problems like work and average. because those problems have real - life context. issues such as zeros in the denominator can be avoided because it doesnt make sense to solve for 0 hours of work or such issues. solutions that cause a zero in the denominator are known as extraneous solutions. lets work an example. solve \\(\frac{2m}{m - 1}-\frac{4m - 4}{m^{2}-1}=1\\). you should get two solutions. do both solutions give you a balanced equation? check each one by substituting the value back into the original equation. do you see any problems? it is important to always check rational equations for extraneous solutions or restrictions on your domain. not every solution that you find works! for each of the given equations, apply your understanding of division to determine the values of \\(x\\) (or other variables) that cannot be solutions to the equation. 1. \\(\frac{5m}{m}+\frac{2m}{m}=1\\) \\(m\
eq0\\) 2. \\(\frac{x^{2}+x - 12}{x - 4}=6\\) \\(x\
eq4\\) 3. \\(\frac{(x + 1)(x + 2)}{x + 1}=\frac{x + 2}{x + 1}\\) \\(x\
eq - 1\\) 4. \\(\frac{3}{2h + 1}=\frac{5}{4h + 3}\\) 5. \\(\frac{5x - 7}{x - 2}=\frac{8}{x - 2}\\) 6. \\(\frac{x - 2}{x + 2}+\frac{3}{x}=\frac{-x}{x + 2}\\) 7. \\(\frac{1}{r}+\frac{1}{r - 3}=\frac{r - 2}{r - 3}\\) 8. \\(\frac{x^{2}}{x + 2}=\frac{4}{x + 2}\\)

Explanation:

Step1: Identify Restrictions (Denominator ≠ 0)

For a rational equation, values that make the denominator zero are not valid solutions (restricted values). We analyze each equation's denominators:

1. Equation: \(\frac{5m}{m} + \frac{2m}{m} = 1\)

Denominators: \(m\).
Restriction: \(m
eq 0\) (since \(m = 0\) makes denominators zero).

2. Equation: \(\frac{x^2 + x - 12}{x - 4} = 6\)

Denominator: \(x - 4\).
Restriction: \(x - 4
eq 0 \implies x
eq 4\).

3. Equation: \(\frac{(x + 1)(x + 2)}{x + 1} = \frac{x + 2}{x + 1}\)

Denominators: \(x + 1\) (both sides).
Restriction: \(x + 1
eq 0 \implies x
eq -1\).

4. Equation: \(\frac{3}{2h + 1} = \frac{5}{4h + 3}\)

Denominators: \(2h + 1\) and \(4h + 3\).

• For \(2h + 1

eq 0 \implies h
eq -\frac{1}{2}\).

• For \(4h + 3

eq 0 \implies h
eq -\frac{3}{4}\).

5. Equation: \(\frac{5x - 7}{x - 2} = \frac{8}{x - 2}\)

Denominator: \(x - 2\) (both sides).
Restriction: \(x - 2
eq 0 \implies x
eq 2\).

6. Equation: \(\frac{x - 2}{x + 2} + \frac{3}{x} = -\frac{x}{x + 2}\)

Denominators: \(x + 2\) and \(x\).

• For \(x + 2

eq 0 \implies x
eq -2\).

• For \(x

eq 0\) (since \(x = 0\) makes the middle denominator zero).

7. Equation: \(\frac{1}{r} + \frac{1}{r - 3} = \frac{r - 2}{r - 3}\)

Denominators: \(r\) and \(r - 3\).

• For \(r

eq 0\) (first denominator).

• For \(r - 3

eq 0 \implies r
eq 3\).

8. Equation: \(\frac{x^2}{x + 2} = \frac{4}{x + 2}\)

Denominator: \(x + 2\) (both sides).
Restriction: \(x + 2
eq 0 \implies x
eq -2\).

Answer:

(Restricted Values):

1. \(m

eq 0\)

1. \(x

eq 4\)

1. \(x

eq -1\)

1. \(h

eq -\frac{1}{2}\), \(h
eq -\frac{3}{4}\)

1. \(x

eq 2\)

1. \(x

eq -2\), \(x
eq 0\)

1. \(r

eq 0\), \(r
eq 3\)

1. \(x

eq -2\)