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Solving One-Step Equations
Choose Course
Algebra 1
One-Variable Equations
Solving One-Step Equations
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Solving One-Step Equations 1.5 - Solution
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Return to Solving One-Step Equations
a
The first step is to substitute
2
2
2
for
y
y
y
in the equation. Then we use
inverse operations
to isolate
x
.
x.
x
.
y
x
=
12
yx=12
y
x
=
1
2
Substitute
y
=
2
y={\color{#0000FF}{2}}
y
=
2
2
x
=
12
{\color{#0000FF}{2}}x=12
2
x
=
1
2
DivEqn
LHS
/
2
=
RHS
/
2
\left.\text{LHS}\middle/2\right.=\left.\text{RHS}\middle/2\right.
LHS
/
2
=
RHS
/
2
x
=
12
2
x=\dfrac{12}{2}
x
=
2
1
2
CalcQuot
Calculate quotient
x
=
6
x=6
x
=
6
b
After substituting
2
2
2
for
y
y
y
we can subtract
2
2
2
from both sides to isolate
x
.
x.
x
.
x
+
y
=
15
x+y=15
x
+
y
=
1
5
Substitute
y
=
2
y={\color{#0000FF}{2}}
y
=
2
x
+
2
=
15
x+{\color{#0000FF}{2}}=15
x
+
2
=
1
5
SubEqn
LHS
−
2
=
RHS
−
2
\text{LHS}-2=\text{RHS}-2
LHS
−
2
=
RHS
−
2
x
=
13
x=13
x
=
1
3
c
Substitute
2
2
2
for
y
y
y
and solve the equation for
x
x
x
by using the
inverse operation
of subtracting
2
,
2,
2
,
meaning, adding
2.
2.
2
.
-
5
=
x
−
y
\text{-}5=x-y
-
5
=
x
−
y
Substitute
y
=
2
y={\color{#0000FF}{2}}
y
=
2
-
5
=
x
−
2
\text{-}5=x-{\color{#0000FF}{2}}
-
5
=
x
−
2
RearrangeEqn
Rearrange equation
x
−
2
=
-
5
x-2=\text{-}5
x
−
2
=
-
5
AddEqn
LHS
+
2
=
RHS
+
2
\text{LHS}+2=\text{RHS}+2
LHS
+
2
=
RHS
+
2
x
=
-
3
x=\text{-}3
x
=
-
3
d
Let's first substitute
2
2
2
for
y
y
y
in the equation.
1
2
x
=
10
\dfrac{1}{{\color{#0000FF}{2}}}x=10
2
1
x
=
1
0
Now we can multiply both sides by
2
2
2
to rewrite find an
equivalent equation
without fractions.
1
2
x
=
10
\dfrac{1}{2}x=10
2
1
x
=
1
0
MultEqn
LHS
⋅
2
=
RHS
⋅
2
\text{LHS} \cdot 2=\text{RHS}\cdot 2
LHS
⋅
2
=
RHS
⋅
2
2
⋅
1
2
x
=
2
⋅
10
2\cdot\dfrac{1}{2}x=2\cdot10
2
⋅
2
1
x
=
2
⋅
1
0
Multiply
Multiply
x
=
20
x=20
x
=
2
0