GRE Prep Club Daily Prep
Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.
Customized
for You
Track
Your Progress
Practice
Pays
Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.
Sep 28Live Online Class with Top GRE Expert Mike Butville
08:00 PM EDT
-10:00 PM EDT
Top scores are possible when you enroll in a powerful GRE course, taught live online + 6 months access to TTP GRE OnDemand video courses included! Class starts Mon/Wed Sept. 28, 2026 - Dec. 9, 2026, 8:00pm-10:00pm EST
If x is different from -2 and also 7
[#permalink]
18 Feb 2026, 14:34
18 Feb 2026, 14:34
Expert Reply
00:00
Question Stats:
99% (01:18) correct
0% (00:00) wrong based on 7 sessions
Hide Show timer Statistics
If $\(x \neq-2, x \neq 7\)$ and $\(\frac{x-3}{x+2}=\frac{x+3}{x-7}\)$, then $\(x=\)$
A. 1
B. 2
C. 3
D. 4
E. 5
A. 1
B. 2
C. 3
D. 4
E. 5
- Part of the project: GRE Daily Quant Challenge - Topic-Wise 2025!!!
- Also replying to the unanswered questions
If x is different from -2 and also 7
[#permalink]
01 Mar 2026, 00:30
01 Mar 2026, 00:30
Expert Reply
Step-by-Step Solution:
1. Start with the given equation:
$$
\(\frac{x-3}{x+2}=\frac{x+3}{x-7}\)
$$
2. Cross-multiply to eliminate the fractions:
$$
\((x-3)(x-7)=(x+3)(x+2)\)
$$
3. Expand both sides (FOIL):
- Left side: $\(x^2-7 x-3 x+21=x^2-10 x+21\)$
- Right side: $\(x^2+2 x+3 x+6=x^2+5 x+6\)$
4. Set the expanded sides equal to each other:
$$
\(x^2-10 x+21=x^2+5 x+6\)
$$
5. Simplify the equation:
Subtract $\(x^2\)$ from both sides:
$$
\(-10 x+21=5 x+6\)
$$
6. Isolate $x$ :
Add $10 x$ to both sides:
$$
\(21=15 x+6\)
$$
Subtract 6 from both sides:
$$
\(15=15 x\)
$$
Divide by 15 :
$$
\(x=1\)
$$
Verification:
Check if $x=1$ is within the defined domain ( $\(x \neq-2, x \neq 7\)$ ): Yes.
Check the values:
- Left side: $\(\frac{1-3}{1+2}=\frac{-2}{3}\)$
- Right side: $\(\frac{1+3}{1-7}=\frac{4}{-6}=-\frac{2}{3}\)$
The values match.
Correct Option:
A. 1
1. Start with the given equation:
$$
\(\frac{x-3}{x+2}=\frac{x+3}{x-7}\)
$$
2. Cross-multiply to eliminate the fractions:
$$
\((x-3)(x-7)=(x+3)(x+2)\)
$$
3. Expand both sides (FOIL):
- Left side: $\(x^2-7 x-3 x+21=x^2-10 x+21\)$
- Right side: $\(x^2+2 x+3 x+6=x^2+5 x+6\)$
4. Set the expanded sides equal to each other:
$$
\(x^2-10 x+21=x^2+5 x+6\)
$$
5. Simplify the equation:
Subtract $\(x^2\)$ from both sides:
$$
\(-10 x+21=5 x+6\)
$$
6. Isolate $x$ :
Add $10 x$ to both sides:
$$
\(21=15 x+6\)
$$
Subtract 6 from both sides:
$$
\(15=15 x\)
$$
Divide by 15 :
$$
\(x=1\)
$$
Verification:
Check if $x=1$ is within the defined domain ( $\(x \neq-2, x \neq 7\)$ ): Yes.
Check the values:
- Left side: $\(\frac{1-3}{1+2}=\frac{-2}{3}\)$
- Right side: $\(\frac{1+3}{1-7}=\frac{4}{-6}=-\frac{2}{3}\)$
The values match.
Correct Option:
A. 1
gmatclubot
If x is different from -2 and also 7 [#permalink]
01 Mar 2026, 00:30
Moderators:
