Transcript 6th Grade Test Prep - Algebra

th
6
Grade Test Prep
Algebra
5.A.2
Translate simple expressions into algebraic expressions
Eighteen less than twice a number
A number squared, plus nineteen
2t - 18
n² + 19
The product of eleven and a number, divided by sixty
11 × k ÷ 60
One hundred more than the quotient of a number
and two hundredths
n ÷ 0.02 + 100
6.A.1
6.A.3
Translate two-step verbal equations and expressions
Joanna receives $8 per hour when she works at the
coffee shop. Last week she earned $256. Write an
equation that can be used to find the number of
hours Joanna worked last week.
Let h = Hours Joanna worked
8h = 256
Write an equation to represent the following. The sum
of 7 and a number, divided by 15, is equal to 33.
Let n = the number
(7 + n) ÷ 15 = 33
6.A.2
5.A.3
Substitute and solve
Evaluate 8a – 2b when a = 7 and b = 4
Evaluate x² + y³ when x = 3 and y = 2
How do you evaluate 6c + d² when c = 9 and d = 5
5.A.4
5.A.5
One step equations
Solve for t:
5t = 25
Step 1: Use the inverse operation (division) to ‘move’ the 5.
5t ÷ 5 = 25 ÷ 5
Step 2: Simplify
t=5
Step 3: Check your work by substituting and solving.
t=5
5t = 25
5(5) = 25
25 = 25
6.A.4
Solve two-step equations
Solve for t:
5 + 4t = 25
Step 1: Start by identifying like terms. Use the inverse operation (subtraction)
to ‘move’ the 5.
5 + (-5) + 4t = 25 + (-5)
4t = 20
Step 2: To isolate the variable, use the inverse operation (division), to ‘move’
the 4.
4t ÷ 4 = 20 ÷ 4
t=5
Step 3: Check your work by substituting and solving.
t=5
5 + 4(5) = 25
5 + 20 = 25
6.A.4
Solve two-step equations
Solve for n:
n
8
- 19 = 7
Step 1: Start with the term that does NOT include the variable. Use the
inverse operation (subtraction) to ‘move’ the 19.
n/8 – 19 (+ 19) = 7 + 19
n/8 = 26
Step 2: To isolate the variable, use the inverse operation (multiplication), to
‘move’ the 8.
n/8 (* 8) = 26 (* 8)
n = 208
Step 3: Check your work by substituting and solving.
n = 208
208 / 8 – 19 = 7 √
6.A.5
Solve simple proportions within context
A car uses 9 gallons of gasoline for a 162-mile drive. How many gallons of gasoline
will the same car use in a 216-mile drive?
Step 1: Write the ratio given in the problem.
The ratio is “162 miles uses 9 gallons.”
As a fraction, the ratio miles/ gallons is 162/9
Step 2: Represent the unknown by a variable.
g = gallons of gasoline for a 216-mile drive
Step 3: Write the ratio and fraction for the unknown variable.
The ratio is “216 miles uses ‘g’ gallons.”
As a fraction, the ratio miles/ gallons is 216 / g
Step 4: Write the ratios as proportions. Cross-multiply to solve.
162g = 1944
g = 12
The car will use 12 gallons of gasoline for a 216-mile drive .
6.A.6
Evaluate formulas for given input values
Julia’s sister drove 224 miles in 4 hours. If Julia’s sister drove at a
constant speed, how fast did she drive? Use the formula d = r t.
d = 224 miles
t = 4 hours
r=?
d=rt
224 = r 4
224 ÷ 4 = r
56 miles per hour = r
Terry just got a $1,200 three-year loan. The loan has a 7%
interest rate. How much interest will Terry pay on the loan?
Use the formula I = P r t.
P = $1,200
r = 7%
t = 3 years
I=?
I=Prt
I = (1200)(.07)(3)
I = 252
Terry will $252 as
interest over the 3
years.
6.A.6
Evaluate formulas for given input values
Which temperature is warmer 34°C or 95°F? Use the formula
°C = 5/9 * (°F – 32)
°C = 34
°F = 95
°C = 5/9 * (°F – 32)
°C = 5/9 * (95 – 32)
°C = 5/9 * (63)
°C = 35°
Thus 95°F = 35°C, which is warmer than 34°C.