2-5 Solving for a Variable
Warm Up
Solve each equation.
1. 5 + x = –2 –7
2. 8m = 43
3.
19
4. 0.3s + 0.6 = 1.5 3
5. 10k – 6 = 9k + 2 8
Holt Algebra 1
2-5 Solving for a Variable
Objectives
Solve a formula for a given variable.
Solve an equation in two or more variables
for one of the variables.
Holt Algebra 1
2-5 Solving for a Variable
Vocabulary
formula
literal equation
Holt Algebra 1
2-5 Solving for a Variable
A formula is an equation that states a rule for a
relationship among quantities.
In the formula d = rt, d is isolated. You can
"rearrange" a formula to isolate any variable by
using inverse operations. This is called solving for a
variable.
Holt Algebra 1
2-5 Solving for a Variable
Solving for a Variable
Step 1 Locate the variable you are asked to
solve for in the equation.
Step 2 Identify the operations on this
variable and the order in which they
are applied.
Step 3 Use inverse operations to undo
operations and isolate the variable.
Holt Algebra 1
2-5 Solving for a Variable
Example 1
Solve the formula d = rt for t
d = rt
Holt Algebra 1
Locate t in the equation.
Since t is multiplied by r, divide both
sides by r to undo the multiplication.
2-5 Solving for a Variable
Example 2
The formula for the area of a triangle is A = bh,
where b is the length of the base, and is the
height. Solve for h.
A=
bh
2A = bh
Locate h in the equation.
Since bh is multiplied by , divide both
sides by to undo the multiplication.
Since h is multiplied by b, divide both
sides by b to undo the multiplication.
Holt Algebra 1
2-5 Solving for a Variable
Remember!
Dividing by a fraction is the same as multiplying
by the reciprocal.
Holt Algebra 1
2-5 Solving for a Variable
Example 3
The formula for a person’s typing speed is
,where s is speed in words per minute,
w is number of words typed, e is number of
errors, and m is number of minutes typing.
Solve for e.
Locate e in the equation.
Since w–10e is divided by m,
multiply both sides by m to
undo the division.
ms = w – 10e
–w –w
ms – w = –10e
Holt Algebra 1
Since w is added to –10e,
subtract w from both sides to
undo the addition.
2-5 Solving for a Variable
Example 3 Continued
The formula for a person’s typing speed is
,where s is speed in words per minute,
w is number of words typed, e is number of
errors, and m is number of minutes typing.
Solve for e.
Since e is multiplied by –10, divide
both sides by –10 to undo the
multiplication.
Holt Algebra 1
2-5 Solving for a Variable
Remember!
Dividing by a fraction is the same as multiplying
by the reciprocal.
Holt Algebra 1
2-5 Solving for a Variable
Example 4
The formula for an object’s final velocity is
f = i – gt, where i is the object’s initial
velocity, g is acceleration due to gravity,
and t is time. Solve for i.
f = i – gt
f = i – gt
+ gt
+gt
f + gt = i
Holt Algebra 1
Locate i in the equation.
Since gt is subtracted from i,
add gt to both sides to undo
the subtraction.
2-5 Solving for a Variable
A formula is a type of literal equation. A
literal equation is an equation with
two or more variables. To solve for one
of the variables, use inverse operations.
Holt Algebra 1
2-5 Solving for a Variable
Example 5
A. Solve x + y = 15 for x.
x + y = 15
–y –y
x
= –y + 15
B. Solve pq = x for q.
pq = x
Holt Algebra 1
Locate x in the equation.
Since y is added to x, subtract y
from both sides to undo the
addition.
Locate q in the equation.
Since q is multiplied by p, divide
both sides by p to undo the
multiplication.
2-5 Solving for a Variable
Example 6
Solve 5 – b = 2t for t.
5 – b = 2t
Locate t in the equation.
Since t is multiplied by 2, divide
both sides by 2 to undo the
multiplication.
Holt Algebra 1
2-5 Solving for a Variable
Example 7
Solve
for V
Locate V in the equation.
VD = m
Since m is divided by V, multiply
both sides by V to undo the
division.
Since V is multiplied by D, divide
both sides by D to undo the
multiplication.
Holt Algebra 1
2-5 Solving for a Variable
Lesson Summary: Part 1
Solve for the indicated variable.
1.
for h
2. P = R – C for C
C=R–P
3. 2x + 7y = 14 for y
4.
for m
5.
for C
Holt Algebra 1
m = x(k – 6 )
C = Rt + S
2-5 Solving for a Variable
Lesson Summary: Part 2
Euler’s formula, V – E + F = 2, relates the
number of vertices V, the number of edges E,
and the number of faces F of a polyhedron.
6. Solve Euler’s formula for F. F = 2 – V + E
7. How many faces does a polyhedron with 8
vertices and 12 edges have? 6
Holt Algebra 1