Question 1

Solve the system. $$\left( \begin{array} { r l l } 
5 x &+ &3 y& - &2 z & = & - 4 \
4 x& -& 7 y &+&7 z & = & 68 \
2 x &-& 5 y &+ &3 z & = & 36
\end{array} \right)$$
A) x = 2, y = -5, z = 2
B) x = 3, y = -3, z = 5
C) x = 2, y = -6, z = 4
D) x = 4, y = -5, z = 4

Accepted Answer

The answer of Solve the system. \[\left( \begin{array} { r...

Question 2

Solve the system. $$\left( \begin{array} { lllll} 
- 4 x &+ &5 y &+& 2 z & =& 0 \
&&4 x &-& 3 z & = & - 24 \
&&&&4 z & = & 16
\end{array} \right)$$

Accepted Answer

The answer of Solve the system. \[\left( \begin{array} { lllll}...

Question 3

A gift store is making a mixture of almonds, pecans, and peanuts, which sell for $3.50 per pound, $5.00 per pound, and $1.00 per pound, respectively. The storekeeper wants to make 15 pounds of the mix to sell at $2.30 per pound. The number of pounds of peanuts is to be three times the number of pounds of pecans. Find the number of pounds of each to be used in the mixture.
A) 6 pounds of almonds, 6 pounds of pecans, and 8 pounds of peanuts
B) 9 pounds of almonds, 2 pounds of pecans, and 9 pounds of peanuts
C) 6 pounds of almonds, 6 pounds of pecans, and 12 pounds of peanuts
D) 3 pounds of almonds, 3 pounds of pecans, and 9 pounds of peanuts

Accepted Answer

The correct mixture must meet the price and weight requirements while adhering to the ratio between peanuts and pecans. Option D correctly balances these factors, achieving the desired total weight and price per pound, with peanuts being three times the amount of pecans.

Question 4

Solve the system. $$\left( \begin{array} { rrrrrr } 
4 x &+ &3 y& -& z & = &- 8 \
&&3 x &-& 5 y & = &12 \
&&4 x &+& 5 y & = & - 19
\end{array} \right)$$
A) x = 2, y = 0, z = -7
B) x = 2, y = -2, z = -7
C) x = -1, y = -3, z = -5
D) x = -1, y = -5, z = -2

Accepted Answer

Substituting x = -1, y = -3, z = -5 into the equations satisfies all three equations.

Question 5

Solve the system. $$\left( \begin{array} { r l } 
5 x& + &5 y &- &3 z & =& - 14 \
&&2 y &+& 4 z & = &20 \
&&3 y &- &2 z & =& 6
\end{array} \right)$$

Accepted Answer

The answer of Solve the system. \[\left( \begin{array} { r...

Question 6

Solve the system. $$\left( \begin{array} { r l } 
3 x - 2 y + 3 z & = 40 \
5 y - 5 z & = - 50 \
4 z & = 20
\end{array} \right)$$

Accepted Answer

The answer of Solve the system. \[\left( \begin{array} { r...

Question 7

Solve the system. $$\left( \begin{array} { r l l } 
5 x &+ &3 y &-& 4 z & = &24 \
&&x & -& 2 z & =& 6 \
&&3 x &+ &4 z & = & - 2
\end{array} \right)$$

Accepted Answer

The answer of Solve the system. \[\left( \begin{array} { r...

Question 8

A small company makes three different types of bird houses. Each type requires the services of three different departments, as indicated by the following table. Type A Type B Type C Cutting department 0.2 hour 0.1 hour 0.3 hour Finishing department 0.5 hour 0.5 hour 0.4 hour Assembly department 0.1 hour 0.1 hour 0.3 hour The cutting, finishing, and assembly departments have available a maximum of 61, 128, and 52 work-hours per week, respectively. How many bird houses of each type should be made per week so that the company is operating at full capacity?
A) 70 houses of type A, 60 houses of type B, and 100 houses of type C
B) 90 houses of type A, 70 houses of type B, and 120 houses of type C
C) 80 houses of type A, 60 houses of type B, and 100 houses of type C
D) 90 houses of type A, 60 houses of type B, and 110 houses of type C

Accepted Answer

Let x, y, and z be the number of Type A, Type B, and Type C bird houses made per week, respectively.

The constraints are:
0.2x + 0.1y + 0.3z ≤ 61
0.5x + 0.5y + 0.4z ≤ 128
0.1x + 0.1y + 0.3z ≤ 52

The objective function is to maximize the total number of bird houses: 
maximize x + y + z

Using a linear programming solver, we get the following optimal solution:
x = 90
y = 70
z = 120

Therefore, the company should make 90 houses of type A, 70 houses of type B, and 120 houses of type C per week to operate at full capacity.

Question 9

Solve the system. $$\left( \begin{array} { c } 
2 x& - &y &+& z& =& - 6 \
3 x& -& 2 y &+& 5 z& =& - 17 \
3 x &+ &y &-& 5 z& = &9
\end{array} \right)$$
A) x = 4, y = 5, z = -1
B) x = 2, y = 4, z = -1
C) x = -1, y = 8, z = 0
D) x = -1, y = 2, z = -2

Accepted Answer

We can solve this system using Gaussian elimination. Adding -3 times the first equation to the second equation gives $$\left( \begin{array} { c } 
2 x & - & y & + & z & = & - 6 \
0 & - & 5 y & + & 13 z & = & 5 \
3 x & + & y & - & 5 z & = & 9
\end{array} \right).$$ Adding -6 times the first equation to the third equation gives $$\left( \begin{array} { c } 
2 x & - & y & + & z & = & - 6 \
0 & - & 5 y & + & 13 z & = & 5 \
0 & + & 7 y & - & 11 z & = & 45
\end{array} \right).$$ Multiplying the second equation by $-2$ and adding it to the third equation gives $$\left(\begin{array}{ccc}2x&-&y&+&z&=&-6\0&-&5y&+&13z&=&5\0&+&17y&-&37z&=&55\end{array}\right).$$ Multiplying the second equation by $-37$ and adding it to the third equation gives $$\left(\begin{array}{ccc}2x&-&y&+&z&=&-6\0&-&5y&+&13z&=&5\0&+&78y&-&510z&=&200\end{array}\right).$$ Multiplying the second equation by 78 and adding it to the third equation gives $$\left(\begin{array}{ccc}2x&-&y&+&z&=&-6\0&-&5y&+&13z&=&5\0&+&53y&-&29z&=&455\end{array}\right).$$ We now have a system of two equations in two unknowns, which we can solve using back substitution. Multiplying the third equation by 5 and adding it to twice the second equation gives $y=2$. Substituting this into the second equation gives $13z=-5-10=-15$, so $z=-1$. Substituting $y=2$ and $z=-1$ into the first equation gives $2x-2-1=-6$, so $x=-1$. Therefore, the answer is $\boxed{\text{(D)}\ x = -1, y = 2, z = -2}$.

Question 10

Solve the system. $$\left( \begin{array} { r l c } 
x + 5 y - 4 z & = & - 10 \
5 y - z & = & 6 \
3 y + 3 z & = & 18
\end{array} \right)$$

Accepted Answer

The answer of Solve the system. \[\left( \begin{array} { r...

Question 11

Solve the system. $\left(\begin{array}{rlll}
2 x&+&5 y&-&3 z & =&-13 \
5 x&-&2 y&+&4 z & =&34 \
&&4 y&+&z & =&-10
\end{array}\right)$
A) x = 0, y = -5, z = 1
B) x = 4, y = -3, z = 2
C) x = 3, y = -6, z = 4
D) x = 0, y = -4, z = 4

Accepted Answer

By solving the system of equations using substitution or elimination, we find that x = 4, y = -3, and z = 2 satisfy all three equations.

Question 12

Solve the system. $\left(\begin{array}{llll}
x&-&2 y&+&z & =&13 \
5 x&+&4 y&-&4 z & =&5 \
-2 x&-&5 y&+&4 z & =&13
\end{array}\right)$
A) x = 5, y = -3, z = 2
B) x = 3, y = -1, z = 5
C) x = 5, y = 0, z = 4
D) x = 5, y = 2, z = 5

Accepted Answer

Solving the system of equations using methods such as substitution, elimination, or matrix operations yields the solution x = 5, y = -3, z = 2.

Question 13

Part of $10,000 is invested at 9%, another part at 11%, and the remainder at 12% yearly interest. The total yearly income from the three investments is $1,100. The sum of the amounts invested at 9% and 11% equals the amount invested at 12%. How much is invested at each rate?
A) $2,500 at 9%, $2,500 at 11%, and $5,000 at 12%
B) $1,500 at 9%, $2,500 at 11%, and $6,000 at 12%
C) $1,500 at 9%, $2,000 at 11%, and $6,000 at 12%
D) $3,000 at 9%, $2,000 at 11%, and $5,000 at 12%

Accepted Answer

The total investment is $10,000, and the total interest earned is $1,100. The sum of the amounts invested at 9% and 11% equals the amount invested at 12%, which means if x is the amount at 12%, then the amount at 9% and 11% combined is also x, making the total investment 2x = $10,000, so x = $5,000. Given the total interest and the rates, the distribution that meets these conditions while also achieving the total interest of $1,100 is $2,500 at 9%, $2,500 at 11%, and $5,000 at 12%.

Question 14

Solve the system. $$\left( \begin{array} { r l } 
x + 2 y - z & = 9 \
x + 3 y + 2 z & = 11 \
- x - 3 y - 2 z & = - 11
\end{array} \right)$$
A) $( 7 k + 5 , - 3 k + 2 , k )$ , $k$ is a real number
B) $( 6 k + 6 , - 2 k + 2 , k )$ , $k$ is a real number
C) $( 7 k + 5 , - 2 k + 2 , k )$ , $k$ is a real number
D) $( 6 k + 6 , - 3 k + 2 , k )$ , $k$ is a real number
E) $( 6 k + 5 , - 3 k + 2 , k )$ , $k$ is a real number

Accepted Answer

The third equation is a multiple of the second, indicating they are not independent, leading to an infinite number of solutions parameterized by $k$. Solving the system with the first two equations and incorporating the parameter for the dependent variable yields the solution in choice A.

Question 15

Solve the system. $\left(\begin{array}{rlll}
4 x&+&2 y&-&z & = & -4 \
4 x&-&4 y&+&4 z & =&40 \
5 x&+&y&-&5 z & = & -14
\end{array}\right)$
A) x = 5, y = -2, z = 2
B) x = 7, y = -4, z = 2
C) x = 2, y = -1, z = 6
D) x = 2, y = -4, z = 4

Accepted Answer

We can solve the system using elimination. Adding the first and third equations gives:
$$9x - 4z = -18$$
Adding twice the first equation to the second equation gives:
$$12x + 4z = 32$$
Now we can eliminate $z$ by adding these two equations:
$$21x = 14$$
$$x = \frac{2}{3}$$
Substituting this into the first equation gives:
$$4\cdot \frac{2}{3} + 2y - z = -4$$
$$\frac{8}{3} + 2y - z = -4$$
$$2y - z = -\frac{20}{3}$$
Substituting our values for $x$ and $z$ into the third equation gives:
$$5\cdot \frac{2}{3} + y - 5\cdot 4 = -14$$
$$y = -\frac{14}{3}$$
Finally, substituting our values for $x$ and $y$ into the first equation gives:
$$4\cdot \frac{2}{3} + 2\cdot\left(-\frac{14}{3}\right) - z = -4$$
$$-\frac{28}{3} - z = -4$$
$$z = 4 - \frac{28}{3} = \frac{4}{3}$$
So the solution is $x = \frac{2}{3}$, $y = -\frac{14}{3}$, and $z = \frac{4}{3}$. We can check that this solution satisfies all three equations. Thus, the answer is choice D.

Question 16

Solve the system. $$\left( \begin{array} { r l l } 
2 x&-&4 y&+&2 z&=&0 \
4 x&+&2 y&-&6 z&=&20 \
5 x&-&3 y&+&6 &z=&-10
\end{array} \right)$$
A) x = 3, y = -2, z = -6
B) x = -1, y = 1, z = -6
C) x = 1, y = -4, z = -4
D) x = 1, y = -1, z = -3

Accepted Answer

Using Gaussian elimination, we obtain the row echelon form of the augmented matrix as follows:
$$\left( \begin{array} { r r r | r } 
2 & -4 & 2 & 0 \
0 & 5 & -7 & 20 \
0 & 0 & 7 & -30
\end{array} \right)$$
From the third equation, $7z=-30$, so $z=-\frac{30}{7}$. Substituting $z$ into the second equation, we get $5x-3y=\frac{194}{7}$. Finally, substituting $z$ and $y$ into the first equation, we get $2x=\frac{56}{7}$, so $x=\frac{28}{7}=4$. Then, we can solve for $y$ using $5x-3y=\frac{194}{7}$, which gives $y=-\frac{34}{7}$. Thus, the solution is $(x,y,z)=(4,-\frac{34}{7},-\frac{30}{7})$, which corresponds to answer choice D.

Question 17

A box contains $6.80 in nickels, dimes, and quarters. There are 40 coins in all, and the sum of the numbers of nickels and dimes is two less than the number of quarters. How many coins of each kind are there?
A) 7 nickels, 12 dimes, and 21 quarters
B) 11 nickels, 11 dimes, and 18 quarters
C) 6 nickels, 13 dimes, and 21 quarters
D) 6 nickels, 10 dimes, and 25 quarters

Accepted Answer

Let's denote the number of nickels, dimes, and quarters as N, D, and Q, respectively. We are given three pieces of information:1. The total value of the coins is $6.80, which translates to the equation: $0.05N + 0.10D + 0.25Q = 6.80$.2. The total number of coins is 40, leading to the equation: $N + D + Q = 40$.3. The sum of the numbers of nickels and dimes is two less than the number of quarters, which gives us: $N + D = Q - 2$.Rearranging the third equation gives us $N + D + 2 = Q$. Substituting this into the second equation, we get $N + D + (N + D + 2) = 40$, simplifying to $2(N + D) + 2 = 40$, and further to $N + D = 19$.Given these equations, we need to find a set of values for N, D, and Q that satisfy all the conditions. Among the options provided, only option A (7 nickels, 12 dimes, and 21 quarters) satisfies all the given conditions:- The total number of coins is $7 + 12 + 21 = 40$.- The sum of the numbers of nickels and dimes is $7 + 12 = 19$, which is indeed two less than the number of quarters (21).- The total value is $0.05*7 + 0.10*12 + 0.25*21 = 0.35 + 1.20 + 5.25 = 6.80$.Therefore, option A is correct.

Question 18

The measure of the largest angle of a triangle is twice the measure of the smallest angle. The sum of the smallest angle and the largest angle is twice the other angle. Find the measure of each angle. A) 55^o, 60^o, 65^o B) 45^o, 60^o, 75^o C) 30^o, 60^o, 90^o D) 40^o, 60^o, 80^o

Accepted Answer

The answer of The measure of the largest angle of...

Question 19

Solve the system. $$\left( \begin{array} { r l } 
3 x - 5 y - 2 z & = 24 \
x - y + z & = 7 \
- 3 x + 4 y - 2 z & = - 24
\end{array} \right)$$
A) x = -1, y = -6, z = 2
B) x = -1, y = -7, z = 2
C) x = -4, y = -3, z = -1
D) x = 2, y = -4, z = 1

Accepted Answer

The answer of Solve the system. \[\left( \begin{array} { r...

Question 20

Solve the system. $\left(\begin{array}{llll}
x&-&5 y&+&2 z & = & 24 \
2 x&+&2 y&+&4 z & = & 12 \
2 x&-&2 y&-&3 z & = & -4
\end{array}\right)$
A) x = 1, y = -6, z = 3
B) x = 0, y = -4, z = 1
C) x = 0, y = -7, z = 1
D) x = 1, y = -3, z = 4

Accepted Answer

To solve the system, we can use elimination method or Gauss-Jordan method. Here, we will use Gauss-Jordan method.
$\left(\begin{array}{rrr|r}
1 & -5 & 2 & 24 \
2 & 2 & 4 & 12 \
2 & -2 & -3 & -4 \end{array}ight)$
$\xrightarrow[	ext{R2}\leftarrow	ext{R2}-2	ext{R1}] {	ext{R3}\leftarrow	ext{R3}-2	ext{R1}}$
$\left(\begin{array}{rrr|r}
1 & -5 & 2 & 24 \
0 & 12 & 0 & -36 \
0 & 8 & -7 & -52 \end{array}ight)$
$\xrightarrow{	ext{R2}\leftarrow\frac{	ext{R2}}{12}}$
$\left(\begin{array}{rrr|r}
1 & -5 & 2 & 24 \
0 & 1 & 0 & -3 \
0 & 8 & -7 & -52 \end{array}ight)$
$\xrightarrow{	ext{R1}\leftarrow	ext{R1}+5	ext{R2}}$
$\left(\begin{array}{rrr|r}
1 & 0 & 2 & 9 \
0 & 1 & 0 & -3 \
0 & 8 & -7 & -52 \end{array}ight)$
$\xrightarrow{	ext{R3}\leftarrow	ext{R3}-8	ext{R2}}$
$\left(\begin{array}{rrr|r}
1 & 0 & 2 & 9 \
0 & 1 & 0 & -3 \
0 & 0 & -7 & 16 \end{array}ight)$
$\xrightarrow{	ext{R3}\leftarrow-\frac{1}{7}	ext{R3}}$
$\left(\begin{array}{rrr|r}
1 & 0 & 2 & 9 \
0 & 1 & 0 & -3 \
0 & 0 & 1 & -\frac{16}{7} \end{array}ight)$
$\xrightarrow{	ext{R1}\leftarrow	ext{R1}-2	ext{R3}}$
$\left(\begin{array}{rrr|r}
1 & 0 & 0 & 1 \
0 & 1 & 0 & -3 \
0 & 0 & 1 & -\frac{16}{7} \end{array}ight)$
Thus, $x=1$, $y=-3$, and $z=\frac{-16}{7}=-2\frac{2}{7}$.
So, the answer is D.

Quiz 15: Systems of Equations: Matrices and Determinants

Solve the system. $\left( \begin{array} { r l l } 5 x &+ &3 y& - &2 z & = & - 4 \\4 x& -& 7 y &+&7 z & = & 68 \\2 x &-& 5 y &+ &3 z & = & 36\end{array} \right)$

Solve the system. $\left( \begin{array} { lllll} - 4 x &+ &5 y &+& 2 z & =& 0 \\&&4 x &-& 3 z & = & - 24 \\&&&&4 z & = & 16\end{array} \right)$

Solve the system. $\left( \begin{array} { rrrrrr } 4 x &+ &3 y& -& z & = &- 8 \\&&3 x &-& 5 y & = &12 \\&&4 x &+& 5 y & = & - 19\end{array} \right)$

Solve the system. $\left( \begin{array} { r l } 5 x& + &5 y &- &3 z & =& - 14 \\&&2 y &+& 4 z & = &20 \\&&3 y &- &2 z & =& 6\end{array} \right)$

Solve the system. $\left( \begin{array} { r l } 3 x - 2 y + 3 z & = 40 \\5 y - 5 z & = - 50 \\4 z & = 20\end{array} \right)$

Solve the system. $\left( \begin{array} { r l l } 5 x &+ &3 y &-& 4 z & = &24 \\&&x & -& 2 z & =& 6 \\&&3 x &+ &4 z & = & - 2\end{array} \right)$

Solve the system. $\left( \begin{array} { c } 2 x& - &y &+& z& =& - 6 \\3 x& -& 2 y &+& 5 z& =& - 17 \\3 x &+ &y &-& 5 z& = &9\end{array} \right)$

Solve the system. $\left( \begin{array} { r l c } x + 5 y - 4 z & = & - 10 \\5 y - z & = & 6 \\3 y + 3 z & = & 18\end{array} \right)$

Solve the system. $\left(\begin{array}{rlll}2 x&+&5 y&-&3 z & =&-13 \\5 x&-&2 y&+&4 z & =&34 \\&&4 y&+&z & =&-10\end{array}\right)$

Solve the system. $\left(\begin{array}{llll}x&-&2 y&+&z & =&13 \\5 x&+&4 y&-&4 z & =&5 \\-2 x&-&5 y&+&4 z & =&13\end{array}\right)$

Part of $10,000 is invested at 9%, another part at 11%, and the remainder at 12% yearly interest. The total yearly income from the three investments is $1,100. The sum of the amounts invested at 9% and 11% equals the amount invested at 12%. How much is invested at each rate?

Solve the system. $\left( \begin{array} { r l } x + 2 y - z & = 9 \\x + 3 y + 2 z & = 11 \\- x - 3 y - 2 z & = - 11\end{array} \right)$

Solve the system. $\left(\begin{array}{rlll}4 x&+&2 y&-&z & = & -4 \\4 x&-&4 y&+&4 z & =&40 \\5 x&+&y&-&5 z & = & -14\end{array}\right)$

Solve the system. $\left( \begin{array} { r l l } 2 x&-&4 y&+&2 z&=&0 \\4 x&+&2 y&-&6 z&=&20 \\5 x&-&3 y&+&6 &z=&-10\end{array} \right)$

A box contains $6.80 in nickels, dimes, and quarters. There are 40 coins in all, and the sum of the numbers of nickels and dimes is two less than the number of quarters. How many coins of each kind are there?

The measure of the largest angle of a triangle is twice the measure of the smallest angle. The sum of the smallest angle and the largest angle is twice the other angle. Find the measure of each angle.

Solve the system. $\left( \begin{array} { r l } 3 x - 5 y - 2 z & = 24 \\x - y + z & = 7 \\- 3 x + 4 y - 2 z & = - 24\end{array} \right)$

Solve the system. $\left(\begin{array}{llll}x&-&5 y&+&2 z & = & 24 \\2 x&+&2 y&+&4 z & = & 12 \\2 x&-&2 y&-&3 z & = & -4\end{array}\right)$

Quiz 15: Systems of Equations: Matrices and Determinants

Solve the system. (5x+3y−2z=−44x−7y+7z=682x−5y+3z=36) \left( \begin{array} { r l l } 5 x &+ &3 y& - &2 z & = & - 4 \\4 x& -& 7 y &+&7 z & = & 68 \\2 x &-& 5 y &+ &3 z & = & 36\end{array} \right) ​5x4x2x​+−−​3y7y5y​−++​2z7z3z​===​−46836​​

Solve the system. (−4x+5y+2z=04x−3z=−244z=16)\left( \begin{array} { lllll} - 4 x &+ &5 y &+& 2 z & =& 0 \\&&4 x &-& 3 z & = & - 24 \\&&&&4 z & = & 16\end{array} \right)​−4x​+​5y4x​+−​2z3z4z​===​0−2416​​

Solve the system. (4x+3y−z=−83x−5y=124x+5y=−19) \left( \begin{array} { rrrrrr } 4 x &+ &3 y& -& z & = &- 8 \\&&3 x &-& 5 y & = &12 \\&&4 x &+& 5 y & = & - 19\end{array} \right) ​4x​+​3y3x4x​−−+​z5y5y​===​−812−19​​

Solve the system. (5x+5y−3z=−142y+4z=203y−2z=6)\left( \begin{array} { r l } 5 x& + &5 y &- &3 z & =& - 14 \\&&2 y &+& 4 z & = &20 \\&&3 y &- &2 z & =& 6\end{array} \right)​5x​+​5y2y3y​−+−​3z4z2z​===​−14206​​

Solve the system. (3x−2y+3z=405y−5z=−504z=20)\left( \begin{array} { r l } 3 x - 2 y + 3 z & = 40 \\5 y - 5 z & = - 50 \\4 z & = 20\end{array} \right)​3x−2y+3z5y−5z4z​=40=−50=20​​

Solve the system. (5x+3y−4z=24x−2z=63x+4z=−2)\left( \begin{array} { r l l } 5 x &+ &3 y &-& 4 z & = &24 \\&&x & -& 2 z & =& 6 \\&&3 x &+ &4 z & = & - 2\end{array} \right)​5x​+​3yx3x​−−+​4z2z4z​===​246−2​​

Solve the system. (2x−y+z=−63x−2y+5z=−173x+y−5z=9) \left( \begin{array} { c } 2 x& - &y &+& z& =& - 6 \\3 x& -& 2 y &+& 5 z& =& - 17 \\3 x &+ &y &-& 5 z& = &9\end{array} \right) ​2x3x3x​−−+​y2yy​++−​z5z5z​===​−6−179​​

Solve the system. (x+5y−4z=−105y−z=63y+3z=18)\left( \begin{array} { r l c } x + 5 y - 4 z & = & - 10 \\5 y - z & = & 6 \\3 y + 3 z & = & 18\end{array} \right)​x+5y−4z5y−z3y+3z​===​−10618​​

Solve the system. (2x+5y−3z=−135x−2y+4z=344y+z=−10) \left(\begin{array}{rlll}2 x&+&5 y&-&3 z & =&-13 \\5 x&-&2 y&+&4 z & =&34 \\&&4 y&+&z & =&-10\end{array}\right) ​2x5x​+−​5y2y4y​−++​3z4zz​===​−1334−10​​

Solve the system. (x−2y+z=135x+4y−4z=5−2x−5y+4z=13) \left(\begin{array}{llll}x&-&2 y&+&z & =&13 \\5 x&+&4 y&-&4 z & =&5 \\-2 x&-&5 y&+&4 z & =&13\end{array}\right) ​x5x−2x​−+−​2y4y5y​+−+​z4z4z​===​13513​​

Part of $10,000 is invested at 9%, another part at 11%, and the remainder at 12% yearly interest. The total yearly income from the three investments is $1,100. The sum of the amounts invested at 9% and 11% equals the amount invested at 12%. How much is invested at each rate?

Solve the system. (x+2y−z=9x+3y+2z=11−x−3y−2z=−11) \left( \begin{array} { r l } x + 2 y - z & = 9 \\x + 3 y + 2 z & = 11 \\- x - 3 y - 2 z & = - 11\end{array} \right) ​x+2y−zx+3y+2z−x−3y−2z​=9=11=−11​​

Solve the system. (4x+2y−z=−44x−4y+4z=405x+y−5z=−14) \left(\begin{array}{rlll}4 x&+&2 y&-&z & = & -4 \\4 x&-&4 y&+&4 z & =&40 \\5 x&+&y&-&5 z & = & -14\end{array}\right) ​4x4x5x​+−+​2y4yy​−+−​z4z5z​===​−440−14​​

Solve the system. (2x−4y+2z=04x+2y−6z=205x−3y+6z=−10) \left( \begin{array} { r l l } 2 x&-&4 y&+&2 z&=&0 \\4 x&+&2 y&-&6 z&=&20 \\5 x&-&3 y&+&6 &z=&-10\end{array} \right) ​2x4x5x​−+−​4y2y3y​+−+​2z6z6​==z=​020−10​​

A box contains $6.80 in nickels, dimes, and quarters. There are 40 coins in all, and the sum of the numbers of nickels and dimes is two less than the number of quarters. How many coins of each kind are there?

The measure of the largest angle of a triangle is twice the measure of the smallest angle. The sum of the smallest angle and the largest angle is twice the other angle. Find the measure of each angle.

Solve the system. (3x−5y−2z=24x−y+z=7−3x+4y−2z=−24) \left( \begin{array} { r l } 3 x - 5 y - 2 z & = 24 \\x - y + z & = 7 \\- 3 x + 4 y - 2 z & = - 24\end{array} \right) ​3x−5y−2zx−y+z−3x+4y−2z​=24=7=−24​​

Solve the system. (x−5y+2z=242x+2y+4z=122x−2y−3z=−4) \left(\begin{array}{llll}x&-&5 y&+&2 z & = & 24 \\2 x&+&2 y&+&4 z & = & 12 \\2 x&-&2 y&-&3 z & = & -4\end{array}\right) ​x2x2x​−+−​5y2y2y​++−​2z4z3z​===​2412−4​​

Solve the system. $\left( \begin{array} { r l l } 5 x &+ &3 y& - &2 z & = & - 4 \\4 x& -& 7 y &+&7 z & = & 68 \\2 x &-& 5 y &+ &3 z & = & 36\end{array} \right)$

Solve the system. $\left( \begin{array} { lllll} - 4 x &+ &5 y &+& 2 z & =& 0 \\&&4 x &-& 3 z & = & - 24 \\&&&&4 z & = & 16\end{array} \right)$

Solve the system. $\left( \begin{array} { rrrrrr } 4 x &+ &3 y& -& z & = &- 8 \\&&3 x &-& 5 y & = &12 \\&&4 x &+& 5 y & = & - 19\end{array} \right)$

Solve the system. $\left( \begin{array} { r l } 5 x& + &5 y &- &3 z & =& - 14 \\&&2 y &+& 4 z & = &20 \\&&3 y &- &2 z & =& 6\end{array} \right)$

Solve the system. $\left( \begin{array} { r l } 3 x - 2 y + 3 z & = 40 \\5 y - 5 z & = - 50 \\4 z & = 20\end{array} \right)$

Solve the system. $\left( \begin{array} { r l l } 5 x &+ &3 y &-& 4 z & = &24 \\&&x & -& 2 z & =& 6 \\&&3 x &+ &4 z & = & - 2\end{array} \right)$

Solve the system. $\left( \begin{array} { c } 2 x& - &y &+& z& =& - 6 \\3 x& -& 2 y &+& 5 z& =& - 17 \\3 x &+ &y &-& 5 z& = &9\end{array} \right)$

Solve the system. $\left( \begin{array} { r l c } x + 5 y - 4 z & = & - 10 \\5 y - z & = & 6 \\3 y + 3 z & = & 18\end{array} \right)$

Solve the system. $\left(\begin{array}{rlll}2 x&+&5 y&-&3 z & =&-13 \\5 x&-&2 y&+&4 z & =&34 \\&&4 y&+&z & =&-10\end{array}\right)$

Solve the system. $\left(\begin{array}{llll}x&-&2 y&+&z & =&13 \\5 x&+&4 y&-&4 z & =&5 \\-2 x&-&5 y&+&4 z & =&13\end{array}\right)$

Solve the system. $\left( \begin{array} { r l } x + 2 y - z & = 9 \\x + 3 y + 2 z & = 11 \\- x - 3 y - 2 z & = - 11\end{array} \right)$

Solve the system. $\left(\begin{array}{rlll}4 x&+&2 y&-&z & = & -4 \\4 x&-&4 y&+&4 z & =&40 \\5 x&+&y&-&5 z & = & -14\end{array}\right)$

Solve the system. $\left( \begin{array} { r l l } 2 x&-&4 y&+&2 z&=&0 \\4 x&+&2 y&-&6 z&=&20 \\5 x&-&3 y&+&6 &z=&-10\end{array} \right)$

Solve the system. $\left( \begin{array} { r l } 3 x - 5 y - 2 z & = 24 \\x - y + z & = 7 \\- 3 x + 4 y - 2 z & = - 24\end{array} \right)$

Solve the system. $\left(\begin{array}{llll}x&-&5 y&+&2 z & = & 24 \\2 x&+&2 y&+&4 z & = & 12 \\2 x&-&2 y&-&3 z & = & -4\end{array}\right)$