Solving by Clearing Fractions #1

To efficiently solve this equation, multiply both sides by:
 $\frac{1}{3}x+4=\frac{5}{2}x-7$  

To efficiently solve this equation, multiply both sides by:
 $\frac{1}{3}x+4=\frac{5}{6}x-7$  

To efficiently solve this equation, multiply both sides by:
 $\frac{1}{2}x+3=\frac{5}{4}x-7$  

To efficiently solve this equation, multiply both sides by:
 $\frac{1}{5}x+3=\frac{1}{4}x-7$  

To efficiently solve this equation, multiply both sides by:
 $\frac{1}{6}x+\frac{3}{2}=\frac{1}{4}x-5$  

Tags

CCSS.5.NF.B.5B

To efficiently solve this equation, multiply both sides by:
 $\frac{2}{3}x-\frac{1}{2}=\frac{4}{5}x-10$  

After multiplying both sides of the equation by 8, which is the resulting equation?
 $\frac{3}{4}x-2=\frac{5}{8}x+3$  

•  $6x-16=5x+24$  

•  $6x-2=5x+3$  

•  $3x-2=5x+3$  

•  $3x-16=5x+24$  

After multiplying both sides of the equation by 10, which is the resulting equation?
 $\frac{3}{2}x+4=\frac{2}{5}x+3$  

•  $15x+40=4x+30$  

•  $15x+4=4x+3$  

•  $30x+4=20x+3$  

•  $3x+40=2x+30$  

After multiplying both sides of the equation by 6, which is the resulting equation?
 $\frac{5}{2}x+3=\frac{7}{6}x+1$  

•  $15x+18=7x+6$  

•  $30x+18=7x+6$  

•  $30x+3=7x+3$  

•  $15x+3=42x+6$  

After multiplying both sides of the equation by 8, which is the resulting equation?
 $\frac{1}{2}x+\frac{3}{8}=\frac{7}{4}x+1$  

•  $4x+3=14x+8$  

•  $4x+3=7x+8$  

•  $8x+3=14x+1$  

•  $8x+3=7x+8$