Question

solve each proportion, then drag it to the correct area of the diagram.

venn diagram with two overlapping circles. left circle labeled ( x = -2 ), right circle labeled ( x = 3 ), overlapping region labeled \both\. below the diagram, five proportion equations:

1. ( x - 3 = \frac{x - 3}{x + 3} )
2. ( x + 3 = \frac{x + 5}{3x + 9} )
3. ( \frac{x - 5}{x} = \frac{x + 13}{x - 3} )
4. ( \frac{x - 3}{x} = \frac{x + 7}{x + 4} )
5. ( 3x + 2 = \frac{2x^2 - 3x - 2}{x - 2} )

Explanation:

Step1: Solve \( x - 3=\frac{x - 3}{x + 3} \)

Multiply both sides by \( x + 3 \) ( \( x
eq - 3 \)): \( (x - 3)(x + 3)=x - 3 \)
\( x^{2}-9=x - 3 \)
\( x^{2}-x - 6 = 0 \)
Factor: \( (x - 3)(x + 2)=0 \)
Solutions: \( x = 3 \) or \( x=-2 \) (check: \( x = 3 \): \( 3-3=\frac{3 - 3}{3 + 3}\Rightarrow0 = 0 \); \( x=-2 \): \( -2-3=\frac{-2 - 3}{-2 + 3}\Rightarrow - 5=-5 \)) → Both

Step2: Solve \( x + 3=\frac{x + 5}{3x + 9} \)

Simplify \( 3x + 9 = 3(x + 3) \), multiply both sides by \( 3(x + 3) \) ( \( x
eq - 3 \)): \( 3(x + 3)^{2}=x + 5 \)
\( 3(x^{2}+6x + 9)=x + 5 \)
\( 3x^{2}+18x + 27=x + 5 \)
\( 3x^{2}+17x + 22 = 0 \)
Discriminant \( \Delta=17^{2}-4\times3\times22=289 - 264 = 25 \)
\( x=\frac{-17\pm5}{6} \), \( x=\frac{-17 + 5}{6}=-2 \), \( x=\frac{-17 - 5}{6}=-\frac{11}{3} \) (check \( x=-2 \): \( -2 + 3=\frac{-2 + 5}{3(-2)+9}\Rightarrow1 = 1 \); \( x =-\frac{11}{3} \) is valid) → \( x=-2 \)

Step3: Solve \( \frac{x - 5}{x}=\frac{x + 13}{x - 3} \)

Cross - multiply: \( (x - 5)(x - 3)=x(x + 13) \)
\( x^{2}-8x + 15=x^{2}+13x \)
\( -21x+15 = 0\Rightarrow x=\frac{15}{21}=\frac{5}{7} \) (not \( -2 \) or \( 3 \))

Step4: Solve \( \frac{x - 3}{x}=\frac{x + 7}{x + 4} \)

Cross - multiply: \( (x - 3)(x + 4)=x(x + 7) \)
\( x^{2}+x-12=x^{2}+7x \)
\( -6x-12 = 0\Rightarrow x=-2 \) (check: \( \frac{-2 - 3}{-2}=\frac{-2 + 7}{-2 + 4}\Rightarrow\frac{5}{2}=\frac{5}{2} \)) → \( x=-2 \)

Step5: Solve \( 3x + 2=\frac{2x^{2}-3x - 2}{x - 2} \)

Multiply both sides by \( x - 2 \) ( \( x
eq2 \)): \( (3x + 2)(x - 2)=2x^{2}-3x - 2 \)
\( 3x^{2}-6x + 2x-4=2x^{2}-3x - 2 \)
\( 3x^{2}-4x-4=2x^{2}-3x - 2 \)
\( x^{2}-x - 2 = 0 \)
Factor: \( (x - 2)(x + 1)=0 \), \( x=-1 \) ( \( x = 2 \) is extraneous) (not \( -2 \) or \( 3 \))

Answer:

• \( x - 3=\frac{x - 3}{x + 3} \): Both
• \( x + 3=\frac{x + 5}{3x + 9} \): \( x=-2 \)
• \( \frac{x - 5}{x}=\frac{x + 13}{x - 3} \): None (not in given categories)
• \( \frac{x - 3}{x}=\frac{x + 7}{x + 4} \): \( x=-2 \)
• \( 3x + 2=\frac{2x^{2}-3x - 2}{x - 2} \): None (not in given categories)

(For the Venn diagram:

• \( x=-2 \) circle: \( x + 3=\frac{x + 5}{3x + 9} \), \( \frac{x - 3}{x}=\frac{x + 7}{x + 4} \)
• Both circle: \( x - 3=\frac{x - 3}{x + 3} \)
• \( x = 3 \) circle: None from these as \( x - 3=\frac{x - 3}{x + 3} \) also has \( x=-2 \))