11.7: Exercise Supplement
- Page ID
- 49415
Exercise Supplement
Solutions by Graphing - Elimination by Addition
For the following problems, solve the systems of equations.
\(\left\{\begin{array}{r}
4x + y = 5\\
-2x + 3y = -13
\end{array}\right.\)
- Answer
-
\((2, -3)\)
\(\left\{\begin{array}{r}
-5x + 2y = 5\\
x + 7y = -1
\end{array}\right.\)
\(\left\{\begin{array}{r}
x - 3y = 17\\
8x + 2y = 46
\end{array}\right.\)
- Answer
-
\((\dfrac{86}{13}, -\dfrac{45}{13})\)
\(\left\{\begin{array}{r}
6m + 5n = -9\\
2m - 4n = 14
\end{array}\right.\)
\(\left\{\begin{array}{r}
3x - 9y = 5\\
-x + 3y = 0
\end{array}\right.\)
- Answer
-
no solution
\(\left\{\begin{array}{r}
y = 2x - 5\\
8x - 75 = 5
\end{array}\right.\)
\(\left\{\begin{array}{r}
x = 8\\
9y = 5x - 76
\end{array}\right.\)
- Answer
-
\((8, -4)\)
\(\left\{\begin{array}{r}
7x - 2y = 4\\
-14x + 4y = -8
\end{array}\right.\)
\(\left\{\begin{array}{r}
y = -x - 7\\
x = y - 5
\end{array}\right.\)
- Answer
-
\((-6, -1)\)
\(\left\{\begin{array}{r}
20x + 15y = -13\\
5x - 20y = 13
\end{array}\right.\)
\(\left\{\begin{array}{r}
x - 6y = 12\\
4x + 6y = 18
\end{array}\right.\)
- Answer
-
\((6, -1)\)
\(\left\{\begin{array}{r}
8x + 9y = 0\\
4x + 3y = 0
\end{array}\right.\)
\(\left\{\begin{array}{r}
-5x + 2y = 1\\
10x - 4y = -2
\end{array}\right.\)
- Answer
-
Dependent (same line)
\(\left\{\begin{array}{r}
2x - 5y = 3\\
5x + 2y = -7
\end{array}\right.\)
\(\left\{\begin{array}{r}
6x + 5y = 14\\
4x - 8y = 32
\end{array}\right.\)
- Answer
-
\((4, -2)\)
\(\left\{\begin{array}{r}
5x - 7y = 4\\
10x - 14y = 1
\end{array}\right.\)
\(\left\{\begin{array}{r}
2m + 10n = 0\\
-4m - 20n = -6
\end{array}\right.\)
- Answer
-
Inconsistent (parallel lines)
\(\left\{\begin{array}{r}
7r - 2s = 6\\
-3r + 5s = -15
\end{array}\right.\)
\(\left\{\begin{array}{r}
28a - 21b = -19\\
21a + 7b = 15
\end{array}\right.\)
- Answer
-
\((\dfrac{2}{7}, \dfrac{9}{7})\)
\(\left\{\begin{array}{r}
72x - 108y = 21\\
18x + 36y = 25
\end{array}\right.\)
Applications
The sum of two numbers is 35. One number is 7 larger than the other. What are the numbers?
- Answer
-
The numbers are 14 and 21.
The difference of two numbers is 48. One number is three times larger than the other. What are the numbers?
A 35 pound mixture of two types of cardboard sells for $30.15. Type I cardboard sells for 90¢ a pound and type II cardboard sells for 75¢ a pound. How many pounds of each type of cardboard were used?
- Answer
-
26 pounds at 90¢; 9 pounds at 75¢
The cost of 34 calculators of two different types is $1139. Type I calculator sells for $35 each and type II sells for $32 each. How many of each type of calculators were used?
A chemistry student needs 46 ml of a 15% salt solution. She has two salt solutions, A and B, to mix together to form the needed 46 ml solution. Salt solution A is 12% salt and salt solution B is 20% salt. How much of each solution should be used?
- Answer
-
\(28 \dfrac{3}{4}\) ml of solution A
\(17 \dfrac{3}{4}\) ml of solution B
A chemist needs 100 ml of a 78% acid solution. He has two acid solutions to mix together to form the needed 100-ml solution. One solution is 50% acid and the other solution is 90% acid. How much of each solution should be used?
One third the sum of two numbers is 12 and half the difference is 14. What are the numbers?
- Answer
-
\(x=32, y=4\)
Two angles are said to be complementary if their measures add to 90°. If one angle measures 8 more than four times the measure of its complement, find the measure of each of the angles.
A chemist needs 4 liters of a 20% acid solution. She has two solutions to mix together to form the 20% solution. One solution is 30% acid and the other solution is 24% acid. Can the chemist form the needed 20% acid solution? If the chemist locates a 14% acid solution, how much would have to be mixed with the 24% acid solution to obtain the needed 20% solution?
- Answer
-
a) no solution
b) 1.6 liters (1600 ml) of the 14% solution
2.4 liters (2400 ml) of the 24% solution.
A chemist needs 80 ml of a 56% salt solution. She has a bottle of 60% salt solution. How much pure water and how much of the 60% salt solution should be mixed to dilute the 60% salt solution to a 56% salt solution?