All Algebra II Resources
Example Questions
Example Question #320 : Solving Equations
Solve for x and y
Solve for x and y by using substitution. Solve the first equation for x.
Next substitute x into the second equation and simplify,
Now use the value for y to solve for x.
Example Question #321 : Solving Equations
Solve the equation.
Solve the equation.
Step 1:
Distribute the on the left side of the equation.
Step 2:
Add to both sides of the equation to isolate the term with the variable.
Step 3:
Divide by both sides of the equation to solve for .
Solution:
Example Question #322 : Solving Equations
Solve the equation.
Solve the equation.
Step 1:
Subtract from both sides of the equation to isolate the term with the variable.
Step 2:
Subtract from both sides of the equation to have only one term with the variable on one side.
Step 3:
Divide by both sides of the equation to solve for .
Solution:
Example Question #453 : Equations
When the sum of five times a number and two is doubled, the result is .
What is the number?
When the sum of five times a number and two is doubled, the result is . What is the number?
Part 1: Writing the equation
In order to find the value of the number, we must take this word problem and convert it into a concrete equation that we can solve. Let's analyze the problem in order to understand what is being asked. In the following problem, the key words will be bolded.
"when the sum of five times a number and two is doubled, the result is "
Sum:
This means that you will have to add. This is written down as the symbol.
Five times a number:
This means that you will multiply a number by . Let's let this unknown number equal . The reason that we are doing this is to make it easier for us to convert the word problem into an actual equation. It is much more simple to calculate "" than it is to calculate "five times a number."
And two:
This means that there is another part to this equation. You must perform an operation to and . As stated in the problem, you must find the sum. This can be written out as .
Is doubled:
This means that you must multiply everything you have written down so far by . The simplest way to write this in algebra is to put in parentheses, and have the outside of the parentheses, as shown below:
The result is :
This means .
We now have all of the information necessary to write a concrete equation from this word problem.
Part 2: Solving the equation
We can solve just as we would for any other equation.
Step 1:
Distribute the two throughout all of the terms inside the parentheses.
Step 2:
Subtract from both sides of the equation.
Step 3:
Divide by both sides of the equation to solve for .
Solution:
Example Question #323 : Solving Equations
Corey is travelling abroad, and is in need of a temporary cell phone plan to avoid roaming charges, so he goes to the nearest cell phone store. One of their price plans includes paying a $100 down payment in addition to fifty cents per hour of talking time. In Corey's budget, he has only $300 to spare for a cell phone plan during his stay.
How many hours can Corey talk on the phone using the cell phone plan until he runs out of money? If necessary, round to the nearest tenth.
Part 1: Writing the equation
The first step to solving this problem is writing down a solvable equation.
In this cell phone plan, you must pay a $100 down payment. In other words, no matter how long or how little you talk on the phone, you must pay this $100. Another part that is added to this $100 is fifty cents per hour of talking time. Because we are looking for the amount of time Corey can talk on the phone, let's let the number of hours. Fifty cents can be represented by . This must be multiplied by to find out how long he can speak on the phone with this plan.
So far, we have .
According to Corey's budget, he only has $300 to spend on this plan. Therefore, we can set this equal to , and solve for .
Part 2: Solving the equation
Subtract 100 on both sides of the equation to isolate the term with the variable .
Divide both sides of the equation by to solve for .
Solution: hours.
Example Question #324 : Solving Equations
Solve for :
In single variable algebraic equations you want to isolate the unknown variable, which is usually a letter, and in this case , then solve for it. It is generally better to do simple operations first (e.g., addition and subtraction), so you aren't left with more complicated fractions.
1) The first step would be to move the 38 to the other side of the equation by subtracting 38 from each side (adding -38).
Giving you:
2) Now to isolate , you would divide each side of the equation by 2.
Giving you:
3) In order to verify the result you can substitute your answer in for and simplify it through operations. If you get the expression 0=0 or any other true expression it is correct:
Example Question #801 : Basic Single Variable Algebra
Solve for x and y
Solve for x and y by using substitution. Solve the first equation for x.
Next substitute x into the second equation and simplify,
Now use the value for y to solve for x.
Example Question #321 : Solving Equations
Solve for x and y.
Solve for x and y by using substitution. Solve the first equation for x.
Next substitute x into the second equation and simplify,
Now use the value for y to solve for x.
Example Question #325 : Solving Equations
Solve for x and y.
Solve for x and y by using substitution. Solve the first equation for x.
Next substitute x into the second equation and simplify,
Now use the value for y to solve for x.
Example Question #322 : Solving Equations
Solve for x and y.
Solve for x and y by using substitution. Solve the first equation for x.
Next substitute x into the second equation and simplify,
Now use the value for y to solve for x.