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The perimeter of a rectangle is 40. The length is 4 more than the width. Find the length and width of the rectangle.
The length is 12 and the width is 8.
The perimeter of a rectangle is 58. The length is 5 more than three times the width. Find the length and width of the rectangle.
The length is 23 and the width is 6.
For [link] we need to remember that the sum of the measures of the angles of a triangle is 180 degrees and that a right triangle has one 90 degree angle.
The measure of one of the small angles of a right triangle is ten more than three times the measure of the other small angle. Find the measures of both angles.
We will draw and label a figure.
| Step 1. Read the problem. |
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| Step 2. Identify what you are looking for. | We are looking for the measures of the angles. |
| Step 3. Name what we are looking for. | Let
the measure of the 1
st angle
the measure of the 2 nd angle |
| Step 4. Translate into a system of equations. | The measure of one of the small angles
of a right triangle is ten more than three times the measure of the other small angle. |
 | |
| The sum of the measures of the angles of
a triangle is 180. | |
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| The system is: |
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Step 5. Solve the system of equations.
We will use substitution since the first equation is solved for a . |
 |
| Substitute 3
b + 10 for
a in the
second equation. |
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| Solve for b . |
 |
 | |
 | |
| Substitute
b = 20 into the first
equation and then solve for a . |
 |
| Step 6. Check the answer in the problem. | We will leave this to you! |
| Step 7. Answer the question. | The measures of the small angles are
20 and 70. |
The measure of one of the small angles of a right triangle is 2 more than 3 times the measure of the other small angle. Find the measure of both angles.
The measure of the angles are 22 degrees and 68 degrees.
The measure of one of the small angles of a right triangle is 18 less than twice the measure of the other small angle. Find the measure of both angles.
The measure of the angles are 36 degrees and 54 degrees.
Heather has been offered two options for her salary as a trainer at the gym. Option A would pay her $25,000 plus $15 for each training session. Option B would pay her $10,000 + $40 for each training session. How many training sessions would make the salary options equal?
| Step 1. Read the problem. | |
| Step 2. Identify what you are looking for. | We are looking for the number of training sessions
that would make the pay equal. |
| Step 3. Name what we are looking for. | Let
Heather’s salary.
the number of training sessions |
| Step 4. Translate into a system of equations. | Option A would pay her $25,000 plus $15
for each training session. |
 | |
| Option B would pay her $10,000 + $40
for each training session | |
 | |
| The system is: |
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|
Step 5. Solve the system of equations.
We will use substitution. |
 |
| Substitute 25,000 + 15 n for s in the second equation. |
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| Solve for n . |
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| Step 6. Check the answer. | Are 600 training sessions a year reasonable?
Are the two options equal when n = 600? |
| Step 7. Answer the question. | The salary options would be equal for 600 training sessions. |
Geraldine has been offered positions by two insurance companies. The first company pays a salary of $12,000 plus a commission of $100 for each policy sold. The second pays a salary of $20,000 plus a commission of $50 for each policy sold. How many policies would need to be sold to make the total pay the same?
There would need to be 160 policies sold to make the total pay the same.
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