3.1 Solving Linear Equations Part I
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3.1 Solving Linear Equations Part I
• A linear equation in one variable can be written in the form: Ax + B = 0
• Linear equations are solved by getting “x” by itself on one side of the equation
• Addition (Subtraction) Property of Equality:
cbca
andcbca
thenbaif
,
3.1 Solving Linear Equations Part I
• Multiplication Property of Equality:
• Since division is the same as multiplying by the reciprocal, you can also divide each side by a number.
• General rule: Whatever you do to one side of the equation, you have to do the same thing to the other side.
cbca
thencbaif
0,
c
b
c
a
thencbaif
0,
3.1 Solving Linear Equations Part I
• Example: Solve by getting x by itself on one side of the equation.
Subtract 7 from both sides:
Divide both sides by 3:
15
453
5273
x
x
x
3.2 Solving Linear Equations Part II- Fractions/Decimals
• As with expressions, you need to combine like terms and use the distributive property in equations.Example:
7
56491058
)715(753
x
x
xx
3.2 Solving Linear Equations Part II- Fractions/Decimals
• Fractions - Multiply each term on both sides by the Least Common Denominator (in this case the LCD = 4):
Multiply by 4:
Reduce Fractions:
Subtract x:
Subtract 5:
17
125
1225
342
45
4
4
32
15
4
1
x
x
xx
xx
xx
3.2 Solving Linear Equations Part II- Fractions/Decimals
• Decimals - Multiply each term on both sides by the smallest power of 10 that gets rid of all the decimals
Multiply by 100:Cancel:Distribute:Subtract 5x:Subtract 50:Divide by 5:
16805
30505
3055010
305510
3.10005.10051.100
3.05.51.
xx
x
xx
xx
xx
xx
3.2 Solving Linear Equations Part II- Fractions/Decimals
• Eliminating fractions makes the calculation simpler:
Multiply by 94:
Cancel:
Distribute:Subtract x:Subtract 10:
104
9410
94102
9452
19494
945
47
94
194
15
47
1
x
x
xx
xx
xx
xx
3.2 Solving Linear Equations Part II
• 1 – Multiply on both sides to get rid of fractions/decimals
• 2 – Use the distributive property• 3 – Combine like terms• 4 – Put variables on one side, numbers on the
other by adding/subtracting on both sides• 5 – Get “x” by itself on one side by multiplying or
dividing on both sides• 6 – Check your answers (if you have time)
3.2 Solving Linear Equations Part II
• Example:Clear fractions:
Combine like terms:
Get variables on one side:
Solve for x:
361
21
32 xxx
1834 xxx
187 xx
186 x3x
3.3 Applications of Linear Equations to General Problems
• 1 – Decide what you are asked to find• 2 – Write down any other pertinent information
(use other variables, draw figures or diagrams )• 3 – Translate the problem into an equation.• 4 – Solve the equation.• 5 – Answer the question posed.• 6 – Check the solution.
3.3 Applications of Linear Equations to General Problems
• Example: The sum of 3 consecutive integers is 126. What are the integers?x = first integer, x + 1 = second integer, x + 2 = third integer
43,42,41
41
1233
12633
126)2()1(
x
x
x
xxx
3.3 Applications of Linear Equations to General Problems
• Example: Renting a car for one day costs $20 plus $.25 per mile. How much would it cost to rent the car for one day if 68 miles are driven?$20 = fixed cost, $.25 68 = variable cost
37$
17$20$
25$.6820$
3.4 Percent Increase/Decrease and Investment Problems
• A number increases from 60 to 81. Find the percent increase in the number.
%35
35.060
21%
216081
increase
increase
3.4 Percent Increase/Decrease and Investment Problems
• A number decreases from 81 to 60. Find the percent increase in the number.
Why is this percent different than the last slide?
%26
26.081
21%
216081
decrease
decrease
3.4 Percent Increase/Decrease and Investment Problems
• A flash drive is on sale for $12 after a 20% discount. What was the original price of the flash drive?
15$
128.
122.0.1
12%20
8.12
x
x
xx
xx
3.4 Percent Increase/Decrease and Investment Problems
• Another Way: A flash drive is on sale for $12 after a 20% discount. What was the original price of the flash drive?Since $12 was on sale for 20% off, it is 100% - 20% = 80% of the original price set up as a proportion (see 3.6):
1580
1200
1200)100(1280100
8012
x
xx
3.4 Percent Increase/Decrease and Investment Problems
• Simple Interest Formula:I = interestP = principalR = rate of interest per yearT = time in years
PRTI
3.4 Percent Increase/Decrease and Investment Problems
• Example: Given an investment of $9500 invested at 12% interest for 1½ years, find the simple interest.
1710$
5.112.09500$
I
PRTI
3.4 Percent Increase/Decrease and Investment Problems
• Example: If money invested at 10% interest for 2 years yields $84, find the principal.
420$2.
84$
2.210.084$
P
PP
PRTI
3.5 Geometry Applications and Solving for a Specific Variable
• A = lw• • P = a + b + c• • •
• Area of rectangle• Area of a triangle• Perimeter of triangle• Sum of angles of a triangle• Area of a circle• Circumference of a circle
321180 mmm2rA rC 2
bhA 21
3.5 Geometry Applications and Solving for a Specific Variable
• Complementary angles – add up to 90
• Supplementary angles – add up to 180
• Vertical angles – the angles opposite each other are congruent
3.5 Geometry Applications and Solving for a Specific Variable
• Find the measure of an angle whose complement is 10 larger.
1. x = degree measure of the angle.
2. 90 – x = measure of its complement
3. 90 – x = 10 + x
4. Subtract 10: Add x: Divide by 2:
40
280
80
x
x
xx
3.6 Applications of Linear Equations to Proportions, d=rt, and Mixture Problems
• Ratio – quotient of two quantities with the same units
Note: percents are ratios where the second number is always 100:
ba
35.%35 10035
3.6 Applications of Linear Equations to Proportions, d=rt, and Mixture Problems
• Percents :
Example: If 70% of the marbles in a bag containing 40 marbles are red, how many of the marbles are red?:# of red marbles =
73.%73 10073
2840)70(.40%70 of
3.6 Applications of Linear Equations to Proportions, d=rt, and Mixture Problems
• Proportion – statement that two ratios are equal:
Solve using cross multiplication:
dc
ba
bcad
3.6 Applications of Linear Equations to Proportions, d=rt, and Mixture Problems
• Solve for x:
Solution:
79
381 x
60
9540
279567
)3(9781
x
x
x
x
3.6 Applications of Linear Equations to Proportions, d=rt, and Mixture Problems
• Example: d=rt (distance = rate time)How long will it take to drive 420 miles at 50 miles per hour?
hourst
t
4.8
50420
50420
3.6 Applications of Linear Equations to Proportions, d=rt, and Mixture Problems
• General form of a mixture problem: x units of an a% solution are mixed with y units of a b% solution to get z units of a c% solution
Equations will always be:
)%()%()%( zcybxa
zyx
3.6 Applications of Linear Equations to Proportions, d=rt, and Mixture Problems
• Example: How many gallons of a 10% indicator solution must be mixed with a 20% indicator solution to get 10 gallons of a 14% solution? Let x = # gallons of 10% solution,
then 10 - x = # gallons of 20% solution :
mixtureofgallonsx
x
xxx
xx
xx
%106
6010
140200101402020010
)10(14)10(2010
)10%(14)10%(20)%(10
3.7 Solving Linear Inequalities in One Variable
• < means “is less than” means “is less than or equal to”
• > means “is greater than” means “is greater than or equal to”
note: the symbol always points to the smaller number
3.7 Solving Linear Inequalities in One Variable
• A linear inequality in one variable can be written in the form:
ax < b (a0)
• Addition property of inequality:
if a < b then a + c < b + c
3.7 Solving Linear Inequalities in One Variable
• Multiplication property of inequality:– If c > 0 then
a < b and ac < bc are equivalent
– If c < 0 thena < b and ac > bc are equivalent
note: the sign of the inequality is reversed when multiplying both sides by a negative number
3.7 Solving Linear Inequalities in One Variable
• Example:36
121
32 xxx
1834 xxx
18 xx182 x
9x-9
3.8 Solving Compound Inequalities
• For any 2 sets A and B, the intersection of A and B is defined as follows:
AB = {x x is an element of A and x is an element of B}
• For any 2 sets A and B, the union of A and B is defined as follows:
AB = {x x is an element of A or x is an element of B}
3.8 Solving Compound Inequalities
• Example:422 and 1013 xx
22 and 93 xx
1 and 3 xx31 x
1 3
3.8 Solving Compound Inequalities
• Example:422or 1013 xx
22or 93 xx
1or 3 xx
1 3