3.E: Solving Linear Systems

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Exercise $\PageIndex{1}$

Determine whether or not the given ordered pair is a solution to the given system.

1. $\left( \frac { 2 } { 3 } , - 4 \right)$;

$\left\{ \begin{array} { l } { 9 x - y = 10 } \\ { 3 x + 4 y = - 14 } \end{array} \right.$

2. $\left( - \frac { 1 } { 2 } , \frac { 3 } { 4 } \right)$;

$\left\{ \begin{array} { l } { 6 x - 8 y = - 9 } \\ { x + 2 y = 1 } \end{array} \right.$

3. $\left( - 5 , - \frac { 7 } { 8 } \right)$;

$\left\{ \begin{array} { l } { x - 16 y = 9 } \\ { 2 x - 8 y = - 17 } \end{array} \right.$

4. $\left( - 1 , \frac { 4 } { 5 } \right)$;

$\left\{ \begin{array} { l } { 2 x + 5 y = 2 } \\ { 3 x - 10 y = - 5 } \end{array} \right.$

Answer

1. Yes

3. No

Exercise $\PageIndex{2}$

Given the graphs, determine the simultaneous solution.

Answer

1. $(-6, 2)$

3. $\varnothing$

Exercise $\PageIndex{3}$

Solve by graphing.

$\left\{ \begin{array} { l } { 2 x + y = 6 } \\ { x - 2 y = 8 } \end{array} \right.$
$\left\{ \begin{array} { l } { 5 x - 2 y = 0 } \\ { x - y = 3 } \end{array} \right.$
$\left\{ \begin{array} { l } { 4 x + 3 y = - 12 } \\ { - 8 x - 6 y = 24 } \end{array} \right.$
$\left\{ \begin{array} { l } { \frac { 1 } { 2 } x + 2 y = 6 } \\ { x + 4 y = - 1 } \end{array} \right.$
$\left\{ \begin{array} { l } { 5 x + 2 y = 30 } \\ { y - 5 = 0 } \end{array} \right.$
$\left\{ \begin{array} { l } { 5 x + 3 y = - 15 } \\ { x + 3 = 0 } \end{array} \right.$
$\left\{ \begin{array} { l } { \frac { 1 } { 3 } x - \frac { 1 } { 2 } y = 2 } \\ { \frac { 1 } { 2 } x + \frac { 3 } { 5 } y = 3 } \end{array} \right.$
$\left\{ \begin{array} { l } { \frac { 2 } { 5 } x + \frac { 1 } { 2 } y = 1 } \\ { \frac { 1 } { 15 } x + \frac { 1 } { 6 } y = - \frac { 1 } { 3 } } \end{array} \right.$

Answer

1. $(4, -2)$

3. $\left( x , - \frac { 4 } { 3 } x - 4 \right)$

5. $(4,5)$

7. $(6,0)$

Exercise $\PageIndex{4}$

Solve by substitution.

$\left\{ \begin{array} { l } { 4 x - y = 12 } \\ { x + 3 y = - 10 } \end{array} \right.$
$\left\{ \begin{array} { l } { 9 x - 2 y = 3 } \\ { x - 3 y = 17 } \end{array} \right.$
$\left\{ \begin{array} { l } { 12 x + y = 7 } \\ { 3 x - 4 y = 6 } \end{array} \right.$
$\left\{ \begin{array} { l } { 3 x - 2 y = 1 } \\ { 2 x + 3 y = - 1 } \end{array} \right.$

Answer

1. $(2,-4)$

3. $\left( \frac { 2 } { 3 } , - 1 \right)$

Exercise $\PageIndex{5}$

Solve by elimination.

$\left\{ \begin{array} { l } { 5 x - 2 y = - 12 } \\ { 4 x + 6 y = - 21 } \end{array} \right.$
$\left\{ \begin{array} { l } { 4 x - 5 y = 12 } \\ { 8 x + 3 y = - 2 } \end{array} \right.$
$\left\{ \begin{array} { l } { 5 x - 3 y = 11 } \\ { 2 x - 4 y = - 4 } \end{array} \right.$
$\left\{ \begin{array} { l } { 7 x + 2 y = 3 } \\ { 3 x + 5 y = - 7 } \end{array} \right.$

Answer

1. $\left( - 3 , - \frac { 3 } { 2 } \right)$

3. $(4,3)$

Exercise $\PageIndex{6}$

Solve using any method.

$\left\{ \begin{array} { l } { 4 x - 8 y = 4 } \\ { x + 2 y = 9 } \end{array} \right.$
$\left\{ \begin{array} { l } { 6 x - 9 y = 8 } \\ { x - y = 1 } \end{array} \right.$
$\left\{ \begin{array} { l } { 2 x - 6 y = - 1 } \\ { 6 x + 10 y = - 3 } \end{array} \right.$
$\left\{ \begin{array} { l } { 2 x - 3 y = 36 } \\ { x - 3 y = 9 } \end{array} \right.$
$\left\{ \begin{array} { l } { 5 x - 3 y = 10 } \\ { - 10 x + 6 y = 3 } \end{array} \right.$
$\left\{ \begin{array} { l } { \frac { 1 } { 2 } x - y = 3 } \\ { 3 x - 6 y = 18 } \end{array} \right.$
$\left\{ \begin{array} { l } { \frac { 3 } { 5 } x - \frac { 1 } { 2 } y = - 1 } \\ { \frac { 1 } { 10 } x + \frac { 3 } { 4 } y = - 1 } \end{array} \right.$
$\left\{ \begin{array} { l } { \frac { 4 } { 3 } x - \frac { 2 } { 5 } y = - \frac { 8 } { 15 } } \\ { \frac { 1 } { 2 } x - \frac { 2 } { 3 } y = - \frac { 11 } { 24 } } \end{array} \right.$

Answer

1. $(5,2)$

3. $\left( - \frac { 1 } { 2 } , 0 \right)$

5. $\varnothing$

7. $\left( - \frac { 5 } { 2 } , - 1 \right)$

Exercise $\PageIndex{7}$

Set up a linear system and solve.

The sum of two integers is $32$. The larger is $4$ less than twice the smaller. Find the integers.
The sum of $2$ times a larger integer and $3$ times a smaller integer is $54$. When twice the smaller integer is subtracted from the larger, the result is $−1$. Find the integers.
The length of a rectangle is $2$ centimeters less than three times its width and the perimeter measures $44$ centimeters. Find the dimensions of the rectangle.
The width of a rectangle is one-third of its length. If the perimeter measures $53 \frac{1}{3}$ centimeters, then find the dimensions of the rectangle.
The sum of a larger integer and $3$ times a smaller is $61$. When twice the smaller integer is subtracted from the larger, the result is $1$. Find the integers.
A total of $$8,600$ was invested in two accounts. One account earned $4 \frac{3}{4}$% annual interest and the other earned $6 \frac{1}{2}$% annual interest. If the total interest for one year was $$431.25$, how much was invested in each account?
A jar consisting of only nickels and dimes contains $76$ coins. If the total value is $$6$, how many of each coin are in the jar?
A nurse wishes to obtain $32$ ounces of a $1.2$% saline solution. How much of a $1$% saline solution must she mix with a $2.6$% saline solution to achieve the desired mixture?
A light aircraft flying with the wind can travel $330$ miles in $2$ hours. The aircraft can fly the same distance against the wind in $3$ hours. Find the speed of the wind.
An executive was able to average $52$ miles per hour to the airport in her car and then board an airplane that averaged $340$ miles per hour. If the total $640$-mile business trip took $4$ hours, how long did she spend on the airplane?

Answer

1. $12, 20$

3. Length: $16$ centimeters; width: $6$ centimeters

5. $12, 25$

7. The jar contains $32$ nickels and $44$ dimes.

9. $27.5$ miles per hour

Exercise $\PageIndex{8}$

Determine whether the given ordered triple is a solution to the given system.

1. $(-2,-1,3)$;

$\left\{ \begin{array} { l } { 4 x - y + 2 z = - 1 } \\ { x - 4 y + 3 z = 11 } \\ { 3 x + 5 y - 4 z = 1 } \end{array} \right.$

2. $(5,-3,-2)$;

$\left\{ \begin{array} { l } { x - 4 y + 6 z = 5 } \\ { 2 x + 5 y - z = - 3 } \\ { 3 x - 4 y + z = 25 } \end{array} \right.$

3. $\left( 1 , - \frac { 3 } { 2 } , - \frac { 4 } { 3 } \right)$;

$\left\{ \begin{array} { l } { 5 x - 4 y + 3 z = 7 } \\ { x + 2 y - 6 z = 6 } \\ { 12 x - 6 y + 6 z = 13 } \end{array} \right.$

4. $\left( \frac { 5 } { 4 } , - \frac { 1 } { 3 } , 2 \right)$;

$\left\{ \begin{array} { c } { 8 x + 9 y + z = 9 } \\ { 4 x + 12 y - 4 z = - 7 } \\ { 12 x - 6 y - z = - 5 } \end{array} \right.$

Answer

1. No

3. Yes

Exercise $\PageIndex{9}$

Solve.

$\left\{ \begin{array} { c } { 2 x + 3 y - z = 1 } \\ { 5 y + 2 z = 12 } \\ { 3 z = 18 } \end{array} \right.$
$\left\{ \begin{array} { c } { 3 x - 5 y - 2 z = 21 } \\ { y - 7 z = 18 } \\ { 4 z = - 12 } \end{array} \right.$
$\left\{ \begin{array} { l } { 4 x - 5 y - z = - 6 } \\ { 3 x + 6 y + 5 z = 3 } \\ { 5 x - 2 y - 3 z = - 17 } \end{array} \right.$
$\left\{ \begin{array} { l } { x - 6 y + 3 z = - 2 } \\ { 5 x + 4 y - 2 z = 24 } \\ { 6 x - 8 y - 5 z = 25 } \end{array} \right.$
$\left\{ \begin{array} { l } { x + 2 y - 2 z = 1 } \\ { 2 x - y - z = - 2 } \\ { 6 x - 3 y - 3 z = 12 } \end{array} \right.$
$\left\{ \begin{array} { l } { 3 x + y + 2 z = - 1 } \\ { 9 x + 3 y + 6 z = - 3 } \\ { 4 x + y + 4 z = - 3 } \end{array} \right.$
$\left\{ \begin{array} { c } { 3 a - 2 b + 5 c = - 3 } \\ { 6 a + 4 b - c = - 2 } \\ { - 6 a + 6 b + 24 c = 7 } \end{array} \right.$
$\left\{ \begin{array} { c } { 9 a - 2 b - 6 c = 10 } \\ { 5 a - 3 b - 10 c = 14 } \\ { - 3 a + 4 b + 12 c = - 20 } \end{array} \right.$

Answer

1. $\left( \frac { 7 } { 2 } , 0,6 \right)$

3. $(-2,-1,3)$

5. $\varnothing$

7. $\left( - \frac { 2 } { 3 } , \frac { 1 } { 2 } , 0 \right)$

Exercise $\PageIndex{10}$

Set up a linear system and solve.

The sum of three integers is $24$. The larger is equal to the sum of the two smaller integers. Three times the smaller is equal to the larger. Find the integers.
The sports center sold $120$ tickets to the Friday night basketball game for a total of $$942$. A general admission ticket cost $$12$, a student ticket cost $$6$, and a child ticket cost $$4$. If the sum of the general admission and student tickets totaled $105$, then how many of each ticket were sold?
A $16$-ounce mixed nut product containing $13.5$% peanuts is to be packaged. The packager has a three-mixed nut product containing $6$%, $10$%, and $50$% peanut concentrations in stock. If the amount of $50$% peanut product is to be one-quarter that of the $10$% peanut product, then how much of each will be needed to produce the desired peanut concentration?
Water is to be mixed with two acid solutions to produce a $25$-ounce solution containing $6$% acid. The acid mixtures on hand contain $10$% and $25$% acid. If the amount of $25$% acid is to be one-half the amount of the $10$% acid solution, how much water will be needed?

Answer

1. $4,8,12$

3. $6$ oz of the $6$% peanut stock, $8$ oz of the $10$% peanut stock, and $2$ oz of the $50$% peanut stock should be mixed.

Exercise $\PageIndex{11}$

Construct the corresponding augmented matrix.

$\left\{ \begin{array} { l } { 9 x - 7 y = 4 } \\ { 3 x - y = - 1 } \end{array} \right.$
$\left\{ \begin{array} { c } { x - 5 y = 12 } \\ { 3 y = - 5 } \end{array} \right.$
$\left\{ \begin{array} { l } { x - y + 2 z = - 6 } \\ { 3 x - 6 y - z = 3 } \\ { - x + y - 5 z = 10 } \end{array} \right.$
$\left\{ \begin{array} { c } { 5 x + 7 y - z = 0 } \\ { - 8 y + z = - 1 } \\ { - x + 3 z = - 9 } \end{array} \right.$

Answer

1. $\left[ \begin{array} { r r | r } { 9 } & { - 7 } & { 4 } \\ { 3 } & { - 1 } & { - 1 } \end{array} \right]$

3. $\left[ \begin{array} { r r r | r } { 1 } & { - 1 } & { 2} &{- 6 } \\ { 3 } & { - 6 } & { - 1} & { 3 } \\ { - 1 } & { 1 } & { - 5 } & { 10 } \end{array} \right]$

Exercise $\PageIndex{12}$

Solve using matrices and Gaussian elimination.

$\left\{ \begin{array} { l } { 4 x + 5 y = 0 } \\ { 2 x - 3 y = 22 } \end{array} \right.$
$\left\{ \begin{array} { l } { 3 x - 8 y = 20 } \\ { 2 x + 5 y = 3 } \end{array} \right.$
$\left\{ \begin{array} { c } { x - y + 4 z = 1 } \\ { - 2 x + 3 y - 2 z = 0 } \\ { x - 6 y + 8 z = 8 } \end{array} \right.$
$\left\{ \begin{array} { l } { - x + 3 y - z = 1 } \\ { 3 x - 6 y + 2 z = - 4 } \\ { 4 x - 3 y + 2 z = - 7 } \end{array} \right.$
$\left\{ \begin{array} { l } { 5 x - 3 y - z = 2 } \\ { x - 6 y + z = 7 } \\ { 2 x + 6 y - 2 z = - 8 } \end{array} \right.$
$\left\{ \begin{array} { l } { x + 2 y + 3 z = 4 } \\ { x + 3 y + z = 3 } \\ { 2 x + 5 y + 4 z = 8 } \end{array} \right.$
$\left\{ \begin{array} { c } { 2 a + 5 b - c = 4 } \\ { 2 a + c = - 2 } \\ { a + b + 3 c = 6 } \end{array} \right.$
$\left\{ \begin{array} { c } { a + 2 b + 3 c = - 7 } \\ { 4 b - 2 c = 8 } \\ { 3 a - c = - 7 } \end{array} \right.$

Answer

1. $(5,-4)$

3. $\left( - 2 , - 1 , \frac { 1 } { 2 } \right)$

5. $\left( x , \frac { 2 } { 3 } x - 1,3 x + 1 \right)$

7. $(-2,2,2)$

Exercise $\PageIndex{13}$

Calculate the determinant.

$\left| \begin{array} { c c} { - 9}&{5 } \\ { - 1}&{3 } \end{array} \right|$
$\left| \begin{array} { c c } { - 5}&{5 } \\ { - 3}&{3 } \end{array} \right|$
$\left| \begin{array} { c c } { 0}&{7 } \\ { 2}&{3 } \end{array} \right|$
$\left| \begin{array} { l l } { 0 } & { b _ { 1 } } \\ { a _ { 2 } }&{b _ { 2 } } \end{array} \right|$
$\left| \begin{array} { r r r } { 2 } & { - 3 } & { 0 } \\ { 1 } & { - 2 } & { - 1 } \\ { 0 } & { 1 } & { 3 } \end{array} \right|$
$\left| \begin{array} { r r r } { 3 } & { 2 } & { - 1 } \\ { 1 } & { - 1 } & { 0 } \\ { 5 } & { - 2 } & { - 4 } \end{array} \right|$
$\left| \begin{array} { r r r } { 5 } & { - 3 } & { - 1 } \\ { 1 } & { - 6 } & { 1 } \\ { 2 } & { 6 } & { - 2 } \end{array} \right|$
$\left| \begin{array} { l l l } { a _ { 1 } } & { 0 } & { 0 } \\ { a _ { 2 } } & { b _ { 2 } } & { 0 } \\ { a _ { 3 } } & { b _ { 3 } } & { c _ { 3 } } \end{array} \right|$

Answer

1. $-22$

3. $-14$

5. $-1$

7. $0$

Exercise $\PageIndex{14}$

Solve using Cramer's rule.

$\left\{ \begin{array} { l } { 2 x - 3 y = - 4 } \\ { 3 x + 5 y = 1 } \end{array} \right.$
$\left\{ \begin{array} { c } { 3 x - y = 2 } \\ { - 2 x + 6 y = 1 } \end{array} \right.$
$\left\{ \begin{array} { l } { 3 x + 5 y = 6 } \\ { 6 x + y = - 6 } \end{array} \right.$
$\left\{ \begin{array} { l } { 6 x - 4 y = - 1 } \\ { - 3 x + 2 y = 2 } \end{array} \right.$
$\left\{ \begin{array} { l } { 5 x + 2 y + 4 z = 4 } \\ { 4 x + 3 y + 2 z = - 5 } \\ { - 5 x - 3 y - 5 z = 0 } \end{array} \right.$
$\left\{ \begin{array} { l } { 2 x - y + 2 z = 1 } \\ { x - 3 y + z = 2 } \\ { 3 x - y - 4 z = - 2 } \end{array} \right.$
$\left\{ \begin{array} { l } { 4 x - y - 2 z = - 7 } \\ { 2 x + y + 6 z = 0 } \\ { 2 x + 2 y + 4 z = - 1 } \end{array} \right.$
$\left\{ \begin{array} { c } { x - y - z = 1 } \\ { 2 x - y + 3 z = 2 } \\ { x + y + z = - 1 } \end{array} \right.$
$\left\{ \begin{array} { l } { 4 x - y + 2 z = - 1 } \\ { 2 x + 3 y - z = 3 } \\ { 6 x + 2 y + z = 2 } \end{array} \right.$
$\left\{ \begin{array} { l } { x - y + 2 z = 1 } \\ { 2 x + 2 y - z = 2 } \\ { 3 x + y + z = 1 } \end{array} \right.$

Answer

1. $\left( - \frac { 17 } { 19 } , \frac { 14 } { 19 } \right)$

3. $\left( - \frac { 4 } { 3 } , 2 \right)$

5. $(2,-5,1)$

7. $\left( - \frac { 3 } { 2 } , 0 , \frac { 1 } { 2 } \right)$

9. $\left( x , - \frac { 8 } { 5 } x + 1 , - \frac { 14 } { 5 } x \right)$

Exercise $\PageIndex{15}$

Determine whether or not the given point is a solution to the system of inequalities.

1. $(-6,1)$;

$\left\{ \begin{array} { l } { - x + y > 2 } \\ { x - 2 y \leq - 1 } \end{array} \right.$

2. $\left( \frac { 1 } { 2 } , - 3 \right)$;

$\left\{ \begin{array} { l } { 4 x - 2 y \geq 8 } \\ { 6 x + 2 y < - 3 } \end{array} \right.$

3. $(-4,-2)$;

$\left\{ \begin{array} { l } { x - y > - 3 } \\ { 2 x + 3 y \leq 0 } \\ { - 3 x + 4 y \geq 4 } \end{array} \right.$

4. $\left( 5 , - \frac { 1 } { 5 } \right)$'

$\left\{ \begin{array} { l } { y < x ^ { 2 } - 25 } \\ { y > \frac { 2 } { 3 } x - 1 } \end{array} \right.$

5. $(-3,-2)$;

$\left\{ \begin{array} { l } { y < ( x - 1 ) ^ { 2 } } \\ { y \leq | x + 1 | - 3 } \end{array} \right.$

6. $\left( 2 , - \frac { 2 } { 3 } \right)$;

$\left\{ \begin{array} { l } { y < 0 } \\ { x ^ { 2 } + y \geq 3 } \end{array} \right.$

Answer

1. Yes

3. Yes

5. Yes

Exercise $\PageIndex{16}$

Graph the solution set.

$\left\{ \begin{array} { l } { y \leq - 4 } \\ { x - 2 y > 8 } \end{array} \right.$
$\left\{ \begin{array} { l } { x + 4 y > 8 } \\ { 2 x - y \leq 4 } \end{array} \right.$
$\left\{ \begin{array} { l } { y - 3 < 0 } \\ { - 2 x + 3 y > - 9 } \\ { x + y \geq 1 } \end{array} \right.$
$\left\{ \begin{array} { l } { y \leq 0 } \\ { 2 x - 6 y < 9 } \\ { - 2 x + 6 y < 9 } \end{array} \right.$
$\left\{ \begin{array} { l } { 2 x + y < 3 } \\ { y > ( x - 2 ) ^ { 2 } - 5 } \end{array} \right.$
$\left\{ \begin{array} { l } { y > | x | } \\ { y \geq - x ^ { 2 } + 6 } \end{array} \right.$
$\left\{ \begin{array} { l } { x - 2 y < 12 } \\ { y \leq ( x - 4 ) ^ { 3 } } \end{array} \right.$
$\left\{ \begin{array} { l } { y + 6 > 0 } \\ { y < \sqrt { x } } \end{array} \right.$

Answer

Sample Exam

Exercise $\PageIndex{17}$

Determine whether or not $(-2, \frac{3}{4})$ is a solution to $\left\{ \begin{array} { l } { 2 x - 8 y = - 10 } \\ { 3 x + 4 y = - 3 } \end{array} \right.$.
Determine whether or not $(−3, 2, −5)$ is a solution to $\left\{ \begin{array} { l } { x - y + 2 z = - 15 } \\ { 2 x - 3 y + z = - 17 } \\ { 3 x + 5 y - 2 z = 10 } \end{array} \right.$.

Answer: 1. Yes

Exercise $\PageIndex{18}$

Solve by graphing.

$\left\{ \begin{array} { l } { x - y = - 5 } \\ { x + y = - 3 } \end{array} \right.$
$\left\{ \begin{array} { l } { 6 x - 8 y = 48 } \\ { \frac { 1 } { 2 } x - \frac { 2 } { 3 } y = 1 } \end{array} \right.$
$\left\{ \begin{array} { l } { \frac { 1 } { 2 } x + y = - 6 } \\ { - 2 x - 4 y = 24 } \end{array} \right.$

Answer

1. $(-4,1)$

3. $\left( x , - \frac { 1 } { 2 } x - 6 \right)$

Exercise $\PageIndex{19}$

Solve by substitution.

$\left\{ \begin{array} { l } { x - 8 y = 10 } \\ { 3 x + 2 y = 17 } \end{array} \right.$
$\left\{ \begin{array} { l } { \frac { 3 } { 2 } x - \frac { 1 } { 6 } y = - \frac { 23 } { 2 } } \\ { \frac { 3 } { 8 } x + \frac { 5 } { 6 } y = - \frac { 11 } { 2 } } \end{array} \right.$
$\left\{ \begin{array} { l } { 5 x - y = 15 } \\ { 2 x - \frac { 2 } { 5 } y = 6 } \end{array} \right.$

Answer: 2. $(-8,-3)$

Exercise $\PageIndex{20}$

Solve.

$\left\{ \begin{array} { l } { 3 x - 5 y = 27 } \\ { 7 x + 2 y = 22 } \end{array} \right.$
$\left\{ \begin{array} { c } { 12 x + 3 y = - 3 } \\ { 5 x + 2 y = 1 } \end{array} \right.$
$\left\{ \begin{array} { l } { 5 x - 3 y = - 1 } \\ { - 15 x + 9 y = 5 } \end{array} \right.$
$\left\{ \begin{array} { l } { 6 a - 3 b + 2 c = 11 } \\ { 2 a - b - 4 c = - 15 } \\ { 4 a - 5 b + 3 c = 23 } \end{array} \right.$
$\left\{ \begin{array} { l } { 4 x + y - 6 z = 8 } \\ { 5 x + 4 y - 2 z = 10 } \\ { 2 x + y - 2 z = 4 } \end{array} \right.$

Answer

1. $(4,-3)$

3. $\varnothing$

5. $\left( x , - x + 2 , \frac { 1 } { 2 } x - 1 \right)$

Exercise $\PageIndex{21}$

Solve using any method.

$\left\{ \begin{array} { c } { x - 5 y + 8 z = 1 } \\ { 2 x + 9 y - 4 z = - 8 } \\ { - 3 x + 11 y + 12 z = 15 } \end{array} \right.$
$\left\{ \begin{array} { l } { 2 x - y + z = 1 } \\ { x - y + 3 z = 2 } \\ { 3 x - 2 y + 4 z = 5 } \end{array} \right.$
$\left\{ \begin{array} { l } { - 5 x + 3 y = 2 } \\ { 4 x + 2 y = - 1 } \end{array} \right.$
$\left\{ \begin{array} { l } { 2 x - 3 y + 2 z = 2 } \\ { x + 2 y - 3 z = 0 } \\ { - x - y + z = - 2 } \end{array} \right.$

Answer

2. $\varnothing$

4. $(2,2,2)$

Exercise $\PageIndex{22}$

Graph the solution set.

$\left\{ \begin{array} { l } { 3 x + 4 y < 24 } \\ { 2 x - 3 y \leq 3 } \\ { y + 1 > 0 } \end{array} \right.$
$\left\{ \begin{array} { l } { x + y < 4 } \\ { y > - ( x + 6 ) ^ { 2 } + 4 } \end{array} \right.$

Answer: 2.

$c08aa597afc28aac062c14cdb629aa50.png$
Figure 3.E.9

Exercise $\PageIndex{23}$

Use algebra to solve the following.

The length of a rectangle is $1$ inch less than twice that of its width. If the perimeter measures $49$ inches, then find the dimensions of the rectangle.
Joe’s $$4,000$ savings is in two accounts. One account earns $3.1$% annual interest and the other earns $4.9$% annual interest. His total interest for the year is $$174.40$. How much does he have in each account?
One solution contains $40$% alcohol and another contains $72$% alcohol. How much of each should be mixed together to obtain $16$ ounces of a $62$% alcohol solution?
Jerry took two buses on the $193$-mile trip to visit his grandmother. The first bus averaged $46$ miles per hour and the second bus was able to average $52$ miles per hour. If the total trip took $4$ hours, then how long was spent in each bus?
A total of $$8,500$ was invested in three interest earning accounts. The interest rates were $2$%, $3$%, and $6$%. If the total simple interest for one year was $$380$ and the amount invested at $6$% was equal to the sum of the amounts in the other two accounts, then how much was invested in each account?
A mechanic wishes to mix $6$ gallons of a $22$% antifreeze solution. In stock he has a $60$% and an $80$% antifreeze concentrate. Water is to be added in the amount that is equal to twice the amount of both concentrates combined. How much water is needed?

Answer

2. Joe has $$1,200$ in the account earning $3.1$% interest and $$2,800$ in the account earning $4.9$% interest

4. Jerry spent $2.5$ hours in the first bus and $1.5$ hours in the second.

6. $4$ gallons of water is needed.

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