An equation consists of two mathematical expressions separated by an equals (=) sign. The expressions may or may not contain variables. Two examples of equations are shown below:

2 + 3 = 5 \qquad\qquad 2 + x = 5

An equation is true if the expression to the left side of the equals sign is equal to the expression on the right side. Otherwise, if the left side is not equal to the right side, then the equation is false. For instance,

  • the equation 2 + 3 = 5 is true, whereas

  • the equation 2+2 = 5 is false.

As a result, the equation

2 + x = 5 is true only if x=3. We say that x=3 is the solution to the equation 2+x=5.

FLAG

If z + 9 = 12, then what is the value of z?

EXPLANATION

If we substitute {\color{blue}{z}}={\color{blue}{3}} into the given equation, we get a true statement:

\begin{align*} {\color{blue}{z}} + 9 &= 12 \\[5pt] {\color{blue}{3}} + 9 &= 12 \\[5pt] 12 &=12 \quad{\color{green}{\checkmark}} \end{align*}

Therefore, the solution is z=3.

FLAG

If $13 + x = 16,$ then $x=$

a
$-3$
b
$3$
c
$1$
d
$10$
e
$6$

If $3 + x = -4$, then $x=$

a
$-3$
b
$1$
c
$-7$
d
$3$
e
$7$

What is the solution to the equation a - 3 = 4?

EXPLANATION

If we substitute {\color{blue}{a}}={\color{blue}{7}} into the given equation, we get a true statement:

\begin{align*} {\color{blue}{a}} - 3 &= 4 \\[5pt] {\color{blue}{7}} - 3 &= 4 \\[5pt] 7 &=7 \quad{\color{green}{\checkmark}} \end{align*}

Therefore, the solution is a=7.

FLAG

If $y - 5 = 2,$ then $y=$

a
$5$
b
$8$
c
$2$
d
$6$
e
$7$

If $12 - y = 4,$ then $y=$

a
$4$
b
$8$
c
$0$
d
$12$
e
$1$

If 2b = 8, then b=

EXPLANATION

If we substitute {\color{blue}{b}}={\color{blue}{4}} into the given equation, we get a true statement:

\begin{align*} 2{\color{blue}{b}} &= 8 \\[5pt] 2\cdot {\color{blue}{4}} &= 8 \\[5pt] 8 &=8 \quad{\color{green}{\checkmark}} \end{align*}

Therefore, the solution is b=4.

FLAG

If $3r = 12,$ then $r=$

a
$1$
b
$-4$
c
$4$
d
$2$
e
$3$

If $-3x = -12,$ then $x=$

a
$3$
b
$4$
c
$-6$
d
$-3$
e
$-4$

If $3z= -24,$ then $z=$

a
$8$
b
$-8$
c
$-9$
d
$-6$
e
$6$

If \dfrac{15}{c} = 5, then c=

EXPLANATION

If we substitute {\color{blue}{c}}={\color{blue}{3}} into the given equation, we get a true statement:

\begin{align*} \dfrac{15}{{\color{blue}{c}}} &= 5 \\[5pt] \dfrac{15}{{\color{blue}{3}}} &= 5 \\[5pt] 5 &= 5 \quad{\color{green}{\checkmark}} \end{align*}

Therefore, the solution is c = \bbox[3pt,Gainsboro]{\color{blue}3}.

FLAG

If $\dfrac{m}{4} = 5,$ then $m=$

a
$5$
b
$-5$
c
$20$
d
$10$
e
$-20$

If $\dfrac{t}{5} = -6,$ then $t=$

a
$t=35$
b
$t=-6$
c
$t=-5$
d
$t=30$
e
$t=-30$
Flag Content
Did you notice an error, or do you simply believe that something could be improved? Please explain below.
SUBMIT
CANCEL