Question

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

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

( 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:

Step 1: 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 \( x=-2 \) (both, since it has both solutions)

Step 2: 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 \( \Delta=17^{2}-4\times3\times22=289 - 264 = 25 \)
\( x=\frac{-17\pm5}{6} \), \( x=\frac{-17 + 5}{6}=-2 \), \( x=\frac{-17 - 5}{6}=-\frac{11}{3} \) (but only \( x=-2 \) is a valid solution here? Wait, re - check:
Wait, original equation: \( x + 3=\frac{x + 5}{3(x + 3)} \)
Multiply both sides by \( 3(x + 3) \): \( 3(x + 3)^{2}=x + 5 \)
\( 3(x^{2}+6x + 9)-x - 5 = 0 \)
\( 3x^{2}+18x+27 - x - 5 = 0 \)
\( 3x^{2}+17x + 22 = 0 \)
As above, \( x=-2 \) (substitute \( x = - 2 \): left side \( -2 + 3=1 \), right side \( \frac{-2 + 5}{3\times(-2)+9}=\frac{3}{3}=1 \); \( x=-\frac{11}{3}\approx - 3.67 \), but \( x =-\frac{11}{3}\), when we check \( 3x + 9=3\times(-\frac{11}{3})+9=-11 + 9=-2
eq0 \), but let's see the solution: when we solve, we get \( x=-2 \) (valid) and \( x =-\frac{11}{3} \) (but the diagram has \( x=-2 \), \( x = 3 \), both. Wait, maybe I made a mistake. Wait, let's try \( x=-2 \):
Left side: \( -2 + 3 = 1 \)
Right side: \( \frac{-2 + 5}{3\times(-2)+9}=\frac{3}{3}=1 \), so \( x=-2 \) is a solution. The other solution \( x =-\frac{11}{3}\) is not in the given solution set of \( x=-2 \), \( x = 3 \), both. So this equation has solution \( x=-2 \) (so \( x=-2 \) circle)

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

Cross - multiply: \( (x + 5)(x + 3)=x(x + 13) \)
\( x^{2}+8x + 15=x^{2}+13x \)
\( - 5x+15 = 0 \)
\( x = 3 \) (so \( x = 3 \) circle)

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

Cross - multiply: \( (x - 3)(x + 4)=x(x + 7) \)
\( x^{2}+x-12=x^{2}+7x \)
\( - 6x-12 = 0 \)
\( x=-2 \) (so \( x=-2 \) circle)

Step 5: 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 \( x=-2 \) (both)

Now, categorize:

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

Answer:

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