Question 111·Hard·Nonlinear Equations in One Variable; Systems in Two Variables

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\frac{x - 4}{x - b} = x - 4

Which of the following are solutions to the given equation, where $b$ is a constant and $b < 3$ ?

I. $b$

II. $4$

III. $b + 1$

A rational equation has a variable in a denominator. First exclude any input that makes the denominator $0$ . Then graph the original sides in Desmos using one allowed value of the constant to spot possible solutions. That graph is only a clue: substitute the choices symbolically to prove which work for every allowed value. A denominator-zero input can never be a solution.

Hints

Hint 1
A rational equation can produce an apparent solution that fails in the original equation. Why should you check a proposed input in the original fraction rather than trust a transformed equation alone?
Hint 2
Choose one value of $b$ that satisfies $b<3$ , such as $0$ , and graph both sides of the original equation. The intersections suggest which inputs to test, but one choice of $b$ cannot prove the answer for every $b$ .
Hint 3
To verify a solution, substitute a proposed value into both sides of the original equation. Check the denominator first, then ask: do the two sides have the same value for every allowed $b$ ?

Step-by-step

Approach 1: Graph a case, then check the choices
Step 1
Use Desmos to spot the candidates
Pick $b=0$ because it satisfies $b<3$ . Type $b=0$ , then graph the original left and right sides as $y=(x-4)/(x-b)$ and $y=x-4$ . Click their intersections: Desmos shows $(1,-3)$ and $(4,0)$ . Because we chose only one value of $b$ , check the proposed inputs symbolically next.
Step 2
Check 4 for every allowed b
Since $b<3$ , the denominator $4-b$ is not $0$ . Substitute $x=4$ into the original equation: $\frac{4-4}{4-b}=0=4-4$ . Both sides are $0$ , so $4$ works.
Step 3
Check the remaining proposed inputs
At $x=b+1$ , the denominator is $(b+1)-b=1$ , so the fraction is defined. Substitute into the original equation: $\frac{(b+1)-4}{(b+1)-b}=\frac{b-3}{1}=b-3=(b+1)-4$ . Both sides match. At $x=b$ , however, the denominator $x-b$ is $0$ , so the fraction is undefined. The solutions are $4$ and $b+1$ , but not $b$ . Choice C.

Approach 2: Factor to find the solutions for any b
Step 1
Clear the denominator
Start with the restriction $x\ne b$ . Multiply both sides by $x-b$ , which is nonzero for an allowed input:
$x-4=(x-4)(x-b).$
Step 2
Put all terms on one side
Subtract $x-4$ from both sides: $0=(x-4)(x-b)-(x-4)$ . The shared factor $x-4$ is now visible.
Step 3
Factor the result
Take out the shared factor $x-4$ : $0=(x-4)(x-b-1)$ . This factored form shows which expressions could equal $0$ .
Step 4
Keep only allowed roots
A product is $0$ if either factor is $0$ , so:
$x-4=0\quad\text{or}\quad x-b-1=0,$
which gives $x=4$ or $x=b+1$ . Since $b<3$ , $4\ne b$ ; also $b+1\ne b$ . Both roots satisfy the original restriction, so the solutions are $4$ and $b+1$ . Choice C.

Question 111·Hard·Nonlinear Equations in One Variable; Systems in Two Variables

Strategy

Hints

Step-by-step

Lessons that teach this

Study Hints

Strategy

Hints

Step-by-step

Lessons that teach this