Solving a linear equation involves isolating the variable on one side of the equation.

This can often be accomplished using the addition principle. The addition principle tells us that adding (or subtracting) the same number on both sides of an equation does not change its solution.

For example, suppose that we want to solve the equation x + 4 = 5.

The variable x is not isolated, because we also have a +4 on the left-hand side.

To isolate x , we can perform the opposite operation, which is subtracting 4 from x. This way, the +4 and -4 will cancel out to 0, which will leave x alone. Remember to do this to both sides of the equation!

\eqalign{ x + 4 &= 5 \\ x+4\,{\color{red}{-}}\,{\color{red}{4}} &=5\,{\color{red}{-}}\,{\color{red}{4}} \\ x+(4-4) &=(5-4) \\ x + 0 &= 1 \\ x &= 1. }

So x=1 is our solution. Let's check that our solution is correct by plugging it back into the original equation:

\eqalign{ x + 4 &= 5 \\ 1 + 4&= 5 \\ 5 &=5 \;{\color{green}{\checkmark}} }

FLAG

Find the solution to the equation 16 + x =10.

EXPLANATION

First, let's rewrite the left-hand-side so that it's similar to what we've seen before.

x + 16 = 10

To isolate x , we need to perform the opposite of adding 16 to x , which is subtracting 16 from x. Remember to perform the same operation on the right-hand side.

\eqalign{ x + 16 &= 10 \\ x + 16 - 16 &= 10 - 16 \\ x + 0 &= -6 \\ x &= -6 }

FLAG

If $z +9 =24,$ then $z=$

a
$-16$
b
$-15$
c
$15$
d
$16$
e
$13$

If $q + 6=7,$ then $q=$

a
$-1$
b
$13$
c
$0$
d
$-13$
e
$1$

Solve the equation x-8 =6.

EXPLANATION

To isolate x , we need to perform the opposite of subtracting 8 from x , which is adding 8 to x. Remember to perform the same operation on the right-hand side.

\eqalign{ x-8 &= 6 \\ x-8+8 &= 6+8 \\ x+0 &= 14 \\ x &= 14 }

FLAG

If $-5 + x = 54$, then $x=$

a
$49$
b
$-39$
c
$-49$
d
$39$
e
$59$

If $x-17=4$, then $x=$

a
$-13$
b
$22$
c
$21$
d
$-21$
e
$13$

What is the solution to the equation -15 = p - 12?

EXPLANATION

First, let's swap the left and right-hand sides so that the variable is on the left-hand side.

p-12 = -15

To isolate p , we need to perform the opposite of subtracting 12 from p , which is adding 12 to p. Remember to perform the same operation on the right-hand side.

\eqalign{ p-12 &= -15\\ p-12 + 12 &= -15 + 12\\ p+0 &=-3\\ p &=-3 }

FLAG

Find the value of $t$ that satisfies the equation $-25=t-11.$

a
$t = 14$
b
$t = -16$
c
$t = -36$
d
$t = -14$
e
$t = 36$

Find the value of $m$ that satisfies the equation $-15=m-10.$

a
$m = 25$
b
$m = -25$
c
$m = -5$
d
$m = 15$
e
$m = 5$

Solve the equation \dfrac{1}{2} = \dfrac{3}{4} + z.

EXPLANATION

First, let's swap the left and right-hand sides so that the variable is on the left-hand side.

\begin{align*} \dfrac{1}{2} &= \dfrac{3}{4} + z\\[5pt] \dfrac{3}{4} + z &= \dfrac{1}{2}\\[5pt] z + \dfrac{3}{4} &=\dfrac{1}{2} \end{align*}

To isolate z , we need to perform the opposite of adding \dfrac{3}{4} to z , which is subtracting \dfrac{3}{4} from z. Remember to perform the same operation on the right-hand side.

\begin{align*} z + \dfrac{3}{4} &=\dfrac{1}{2}\\[5pt] z+ \dfrac{3}{4} - \dfrac{3}{4} &=\dfrac{1}{2} - \dfrac{3}{4}\\[5pt] z + 0 &=\dfrac{2}{4} - \dfrac{3}{4}\\[5pt] z &=\dfrac{2-3}{4}\\[5pt] z &=-\dfrac{1}{4} \end{align*}

FLAG

Find the solution to the equation $1.8=w-4.7.$

a
$w = 2.9$
b
$w = -2.9$
c
$w = -6.5$
d
$w = 3.9$
e
$w = 6.5$

Solve the equation $\dfrac{5}{6} = z - \dfrac{1}{3}.$

a
$z = \dfrac{1}{2}$
b
$z = \dfrac{5}{4}$
c
$z = \dfrac{7}{6}$
d
$z = \dfrac{1}{6}$
e
$z = \dfrac{6}{5}$
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