One-Variable Linear Equations and Inequalities

 $x+9=18-2x$  Find the value of x.

• x = 5

• x = 2

• x = 3

• x = 4

 $4(2x-3)=6x+2$  

Solve the following equation for x.

• x = 5/2

• x = 7

• x = 1

• no solution

Tags

CCSS.8.EE.C.7B

Which of the following equations would have no solution?

• $3x-2=2-3x$

• $10+4x=4x+10$
  

• $9+2x=3x+9$
  

• $12x+4=12x+8$
  

Tags

CCSS.8.EE.C.7A

Which of the following equations would have infinite solutions?

• $3x+9=2x+9$

• $8x=8x+4$
  

• $3(4x+4)=12x+4$
  

• $-5x+10=10-5x$
  

Tags

CCSS.8.EE.C.7A

 $\frac{x+2}{8}=\frac{3}{4}$  Solve.

• x = 2

• x = 6

• x = 4

• x = 8

Tags

CCSS.7.EE.B.4A

What error did the student make when solving the following equation:
 $\frac{1}{5}x+6=\frac{3}{5}x+5$  
 $1x+6=3x+5$  
 $-2x+6=5$  
 $-2x=-1$  
 $x=1/2$  

• The student found an incorrect common denominator

• The student did not multiply all terms by the common denominator

• The student did not use opposite operations appropriately

• The student did not correctly eliminate the variable from one side of the equation

What is the LCM (Least Common Multiple) that you would multiply everything by to get rid of the fractions?

 $\frac{3}{4}\left(x-4\right)\ +\ 3=\frac{1}{8}\left(x-16\right)\$  

• 2

• 4

• 8

• 12

What would the next step look like after you get rid of the fractions  by multiplying everything by the LCM?

 $\frac{3}{4}\left(x-4\right)\ +\ 3=\frac{1}{8}\left(x-16\right)\$  

•  $3\left(x-4\right)\ +12=2\left(x\ -\ 16\right)$  

•  $6\left(x-4\right)\ +\ 24=x\ -16$  

•  $x-4+3=x-16$  
  

•  $6\left(x\ -4\right)\ +\ 3\ =\ x\ -\ 16$  

What would be the solution for this equation?

 $\frac{3}{4}\left(x-4\right)\ +\ 3=\frac{1}{8}\left(x-16\right)\$  

•  $x=-32$  

•  $x=-\frac{16}{5}$  

• The equation has no solution.

•  $x=1$  

Tags

CCSS.8.EE.C.7B

Which is a solution of

 $5(v+3)-10v>-10$  

• v < 5

• v > 5

• v > -5

• v < -5