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Solving Literal Equations
Choose Course
Algebra 1
One-Variable Equations
Solving Literal Equations
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Exercise 1.4 - Solution
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Return to Solving Literal Equations
a
Let's begin by solving the equation for
y
by subtracting
2
x
from both sides.
y
+
2
x
=
5
⇔
y
=
5
−
2
x
To find the value of
y
when
x
=
-
1
,
we substitute
-
1
for
x
in the equation.
y
=
5
−
2
x
Substitute
x
=
-
1
y
=
5
−
2
(
-
1
)
MultNegNegOnePar
-
a
(
-
b
)
=
a
⋅
b
y
=
5
+
2
AddTerms
Add terms
y
=
7
b
First, we'll solve the equation for
y
with inverse operations.
4
x
=
3
y
−
7
AddEqn
LHS
+
7
=
RHS
+
7
4
x
+
7
=
3
y
DivEqn
LHS
/
3
=
RHS
/
3
3
4
x
+
7
=
y
RearrangeEqn
Rearrange equation
y
=
3
4
x
+
7
To find the value of
y
when
x
=
-
1
we'll substitute
x
with
-
1
.
y
=
3
4
x
+
7
Substitute
x
=
-
1
y
=
3
4
(
-
1
)
+
7
MultPosNeg
a
(
-
b
)
=
-
a
⋅
b
y
=
3
-
4
+
7
AddTerms
Add terms
y
=
3
3
CalcQuot
Calculate quotient
y
=
1