All Algebra II Resources
Example Questions
Example Question #3 : Linear Systems With Two Variables
What is a solution to this system of equations?
Substitute equation 2. into equation 1.,
so,
Substitute into equation 2:
so, the solution is .
Example Question #1 : Systems Of Equations
A man in a canoe travels upstream 400 meters in 2 hours. In the same canoe, that man travels downstream 600 meters in 2 hours.
What is the speed of the current, , and what is the speed of the boat in still water, ?
More information is needed
This problem is a system of equations, and uses the equation .
Start by assigning variables. Let stand for the rate of the boat, let stand for the rate of the current.
When the boat is going upstream, the total rate is equal to . You must subtract because the rates are working against each other—the boat is going slower than it would because it has to work against the current.
Using our upstream distance (400m) and time (2hr) from the question, we can set up our rate equation:
When the boat is going downstream, the total rate is equal to because the boat and current are working with each other, causing the boat to travel faster.
We can refer to the downstream distance (600m) and time (2hr) to set up the second equation:
From here, use elimination to solve for and .
1. Set up the system of equations, and solve for .
2. Subsitute into one of the equations to solve for .
Example Question #2331 : Algebra Ii
The numerator of a fraction is and the quotient is . What is the denominator?
Step 1: set up the equation.
Step 2: cross-multiply and rewrite the equation so that is by itself.
Step 3: Simplify the fraction to find .
Example Question #2332 : Algebra Ii
Solve:
Step 1: Multiply each side by the least common multiple (LCM) of the two demonominators. The LCM is twelve.
Step 2: Reduce the fractions.
Step 3: Solve for .
Example Question #14 : Solving Equations
Solve the equation for :
Step 1: Represent the entire right side of the equation as a fraction, using 6 as a common denominator.
Step 2: Cross multiply and solve for .
Example Question #15 : Solving Equations
Two cars are 1900 miles apart. Car 1 leaves at 4pm and is travelling east at 40mph. Car 2 leaves an hour later, and is travelling west at 50mph.
How long until the cars meet?
19 hours
Not enough information given
21 hours 30 min
21 hours and 10 min
21 hours and 40 min
21 hours and 40 min
Step 1: Set up your equation
Consider the rate and driving time of each car. Additionally, Car 2 leaves an hour after car one, it is necessary to adjust for this in the equation.
Determine the variables:
Let = the rate of Car 1
Let = the rate of Car 2
Let = the amount of time after Car 1 leaves
Therefore:
Step 2: Solve
Example Question #491 : Basic Single Variable Algebra
Solve for x:
Solve for x:
Subtract from each side:
Divide by
Example Question #2333 : Algebra Ii
Solve this sytem of equations:
How many solutions are there?
Not enough information is given
There is an infinte number of solutions
There are exactly 2 solutions
There is only 1 solution
There is no solution
There is an infinte number of solutions
Write the second equation in slope-intercept form
This is the same as the first equation.
The two equations represent the same line, therefore there are an infinite number of solutions.
Example Question #18 : Solving Equations
Three brothers set out to paint a greenhouse together. Tom can paint the greenhouse in 9 hours, working alone; Dick can paint it in 7 hours; Harry can paint it in 11 hours. If the brothers start painting at at 9:00 AM and do not stop for a break, which of the following times will come closest to the time at which they finish?
Let be the number of hours it takes to paint the greenhouse together.
Work problems can be solved by looking at them as rate problems.
Since Tom, Dick, and Harry would take 9, 7, and 11 hours, respectively, to finish the job, the rates at which each works can be viewed as follows:
Tom: greenhouse per hour; greenhouse painted.
Dick: greenhouse per hour; greenhouse painted.
Harry: greenhouse per hour; greenhouse painted.
Since their work adds up to one greenhouse painted, We can set up, and solve for in, the equation:
Using decimal approximations:
hours
Since 0.90 hours is approximately equal to minutes, it will take 2 hours 54 minutes to finish painting the greenhouse - that is, it will be 11:54 AM when they stop. Of the given choices, 12:00 noon comes closest.
Example Question #19 : Solving Equations
A large tub has two faucets. The hot water faucet, if turned all the way up, can fill the tub in 16 minutes; the cold water faucet, in 12 minutes. Which of the following choices comes closest to the amount of time it takes for both faucets working together to fill the tub?
Work problems can be solved by looking at them as rate problems.
The hot faucet can fill up the tub at a rate of 16 minutes per tub, or tub per minute. The cold faucet, similarly, can fill up the tub at a rate of 12 minutes per tub, or tub per minute.
Suppose the tub fills up in minutes. Then, at the end of this time, the hot faucet has filled up tub, and the cold faucet has filled up tub, for a total of one tub. We can set up this equation and solve for :
Of the given choices, 7 minutes comes closest.
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