Question

solve each proportion, then drag it to the correct area of the diagram.
$x = -2$ $\text{both}$ $x = 3$
$x - 3 = \frac{x - 3}{x + 3}$ $x + 3 = \frac{x + 5}{3x + 9}$ $\frac{x + 5}{x} = \frac{x + 13}{x + 3}$ $\frac{x - 3}{x} = \frac{x + 7}{x + 4}$ $3x + 2 = \frac{2x^2 - 3x + 2}{x - 2}$

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: \( x=-3 \) is excluded, both \( x = 3 \) and \( x=-2 \) work) → 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 \( \Delta=17^{2}-4\times3\times22=289 - 264 = 25 \)
\( x=\frac{-17\pm5}{6} \), \( x=\frac{-12}{6}=-2 \) or \( x=\frac{-22}{6}=-\frac{11}{3} \) (check \( x=-2 \) works, \( x =-\frac{11}{3}\) works? Wait, re - check: original equation, when \( x=-2 \), left \( -2 + 3 = 1 \), right \( \frac{-2 + 5}{3\times(-2)+9}=\frac{3}{3}=1 \). When \( x =-\frac{11}{3} \), \( 3x+9=3\times(-\frac{11}{3})+9=-11 + 9=-2
eq0 \), but let's re - solve. Wait, maybe miscalculation. Wait, \( x + 3=\frac{x + 5}{3(x + 3)} \), multiply both sides by \( 3(x + 3) \): \( 3(x + 3)^{2}=x + 5 \), \( 3x^{2}+18x + 27-x - 5 = 0 \), \( 3x^{2}+17x + 22 = 0 \), roots \( x=\frac{-17\pm\sqrt{289 - 264}}{6}=\frac{-17\pm5}{6} \), so \( x=-2 \) or \( x =-\frac{11}{3} \). But when \( x=-2 \), it works. Let's check the problem's target (maybe I made a mistake). Wait, maybe the problem expects us to see that when we simplify, \( 3x + 9 = 3(x + 3) \), so the equation is \( x + 3=\frac{x + 5}{3(x + 3)} \), cross - multiply: \( 3(x + 3)^{2}=x + 5 \), \( 3x^{2}+18x + 27=x + 5 \), \( 3x^{2}+17x + 22 = 0 \). But maybe the intended solution: let's try \( x=-2 \): left \( -2 + 3 = 1 \), right \( \frac{-2 + 5}{3\times(-2)+9}=\frac{3}{3}=1 \), so \( x=-2 \) is a solution. Any other? \( x =-\frac{11}{3}\approx - 3.666 \), then \( x + 3=-\frac{2}{3} \), \( 3x + 9=-2 \), right \( \frac{-\frac{11}{3}+5}{-2}=\frac{\frac{4}{3}}{-2}=-\frac{2}{3} \), so it also works. But the problem's Venn has \( x=-2 \), \( x = 3 \), both. Maybe I misread the equation. Wait, the second equation is \( x + 3=\frac{x + 5}{3x + 9} \), maybe \( 3x + 9 \) is \( 3x+9 \), but maybe it's a typo? Wait, maybe the equation is \( x + 3=\frac{x + 5}{3x + 9} \), and we can see that when \( x=-2 \), it works. Let's assume the problem expects \( x=-2 \) (maybe the other root is extraneous? No, it's not. But maybe in the problem's context, we consider \( x=-2 \)) → \( x=-2 \)

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

Cross - multiply: \( (x + 5)(x + 3)=x(x + 13) \)
\( x^{2}+8x + 15=x^{2}+13x \)
\( - 5x+15 = 0 \)
\( x = 3 \) (check: \( x
eq0,-3 \), \( x = 3 \) works) → \( x = 3 \)

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

Cross - multiply: \( (x - 3)(x + 4)=x(x + 7) \)
\( x^{2}+x-12=x^{2}+7x \)
\( - 6x-12 = 0 \)
\( x=-2 \) (check: \( x
eq0,-4 \), \( x=-2 \) works) → \( x=-2 \)

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

Multiply both sides by \( x - 2 \) ( \( x
eq2 \)): \( (3x + 2)(x - 2)=2x^{2}-3x + 2 \)
\( 3x^{2}-6x+2x - 4=2x^{2}-3x + 2 \)
\( 3x^{2}-4x - 4-2x^{2}+3x - 2 = 0 \)
\( x^{2}-x - 6 = 0 \)
Factor: \( (x - 3)(x + 2)=0 \)
Solutions: \( x = 3 \) or \( x=-2 \) (check: \( x = 2 \) is excluded, both \( x = 3 \) and \( x=-2 \) work) → Both

Answer:

• \( x - 3=\frac{x - 3}{x + 3} \): Both
• \( x + 3=\frac{x + 5}{3x + 9} \): \( x=-2 \)
• \( \frac{x + 5}{x}=\frac{x + 13}{x + 3} \): \( x = 3 \)
• \( \frac{x - 3}{x}=\frac{x + 7}{x + 4} \): \( x=-2 \)
• \( 3x + 2=\frac{2x^{2}-3x + 2}{x - 2} \): Both

(To place in the Venn:

• \( x=-2 \) circle: \( x + 3=\frac{x + 5}{3x + 9} \), \( \frac{x - 3}{x}=\frac{x + 7}{x + 4} \)
• \( x = 3 \) circle: \( \frac{x + 5}{x}=\frac{x + 13}{x + 3} \)
• Both circle: \( x - 3=\frac{x - 3}{x + 3} \), \( 3x + 2=\frac{2x^{2}-3x + 2}{x - 2} \))