Question 1

Solve the system of linear equations by the method of elimination.Find (x,y)and check your solution algebraically.​ $$\left\{ \begin{array} { l } 
0.2 x - 0.5 y = - 27.1 \
0.3 x + 0.4 y = 70.9
\end{array} \right.$$ ​

Accepted Answer

By multiplying the first equation by 3 and the second by 2, we can eliminate y. Solving the resulting equations gives x = 107 and y = 97.

Question 2

Solve the system of linear equations by the method of elimination.Find (x,y)and check your solution algebraically.​ $$\left\{ \begin{array} { l } 
4 b + 4 m = \frac { 156 } { 35 } \\
4 b + 11 m = \frac { 471 } { 35 }
\end{array} \right.$$ ​

Accepted Answer

The answer of Solve the system of linear equations by...

Question 3

Seven hundred gallons of 87-octane gasoline is obtained by mixing 87-octane gasoline with 90-octane gasoline.Use a graphing utility to graph the two given equations.Let x and y represent the number of gallons of the 87-octane gasoline and 90-octane gasoline.As the number of gallons of the 87-octane gasoline increases,how does the number of gallons of the 90-octane gasoline change?​ $$\left\{ \begin{array} { l } 
x + y = 700 \
87 x + 90 y = 60,900
\end{array} \right.$$ ​

Accepted Answer

The answer of Seven hundred gallons of 87-octane gasoline is...

Question 4

Solve the system of linear equations by the method of elimination.Find (x,y)and check your solution algebraically. ​​ $$\left\{ \begin{array} { c } 
5 x + 7 y = 21 \
- 14 x - 18 y = - 38
\end{array} \right.$$ ​

Accepted Answer

To solve the system of equations by elimination, we first aim to eliminate one of the variables by making the coefficients of either x or y the same in both equations. We can do this by multiplying the first equation by 2 and the second equation by 1 (to simplify calculations, we notice that multiplying the first equation by 2 will make the coefficients of y equal but opposite in sign, facilitating their elimination):$$2(5x + 7y) = 2(21)$$$$2(5x) + 2(7y) = 42$$$$10x + 14y = 42$$The second equation remains unchanged:$$-14x - 18y = -38$$Now, we add the two equations together to eliminate y:$$10x + 14y - 14x - 18y = 42 - 38$$$$-4x - 4y = 4$$$$-4(x + y) = 4$$$$x + y = -1$$ (Equation 1)To find the value of one of the variables, we can substitute the value of x from one of the original equations or find another combination. Let's use the first original equation to express x in terms of y:$$5x + 7y = 21$$$$5x = 21 - 7y$$$$x = \frac{21 - 7y}{5}$$Now, we substitute $x + y = -1$ to solve for y:$$\frac{21 - 7y}{5} + y = -1$$$$21 - 7y + 5y = -5$$$$-2y = -26$$$$y = 13$$Substituting y = 13 into $x + y = -1$:$$x + 13 = -1$$$$x = -14$$Therefore, the solution is $(-14, 13)$, which corresponds to choice D.

Question 5

Use any method to solve the system of linear equations,find (x,y). ​​ $$\left\{ \begin{array} { c } 
4 x - 3 y = 16 \
- 5 x + 3 y = - 23
\end{array} \right.$$ ​

Accepted Answer

Adding the two equations directly cancels out $y$, leading to $-x = -7$, so $x = 7$. Substituting $x = 7$ into the first equation gives $4(7) - 3y = 16$, solving for $y$ yields $y = 4$.

Question 6

Solve the system of linear equations by the method of elimination.Find (x,y)and check your solution algebraically.​ $$\left\{ \begin{array} { c } 
5 x + 3 y = 6 \
3 x - y = 5
\end{array} \right.$$ ​

Accepted Answer

We can solve the system of equations by eliminating y first. To eliminate y, we need to multiply the first equation by 2 and subtract the second equation from it:

2x - 4y = -10
-2x + 6y = 6

Adding these two equations, we get:

2y = -4

Solving for y, we get:

y = -2

Substituting y = -2 in either of the original equations, we get:

3x + 2(-2) = -4

Simplifying this equation, we get:

3x - 4 = -4

Solving for x, we get:

x = 0

Therefore, the solution to the system of equations is (0,-2). We can check our solution by substituting x = 0 and y = -2 in both equations:

3(0) + 2(-2) = -4

-2(0) + 3(-2) = -6

Both equations are true, so our solution is correct.

Out of the given options, the only choice with the solution (0,-2) is E.

Question 7

​Solve the system of linear equations by the method of elimination,find (x,y).​ $$\left\{ \begin{array} { c } 
- 5 x + 2 y = - 38 \
3 x - y = 23
\end{array} \right.$$ ​ ​

Accepted Answer

By adding the given equations, we get:
-5x + 9y = -12
Multiplying the second equation by 2, we get:
-6x + 6y = 6
Multiplying the first equation by 2 and adding it to the second equation, we get:
3y = 18
y = 6
Substituting y = 6 in the first equation, we get:
-5x + 54 = -12
-5x = -66
x = 13.2
So, the solution is (13.2, 6). Therefore, the correct answer is option B.

Question 8

Solve the system of linear equations by the method of elimination.Find (x,y)and check your solution algebraically.​ $$\left\{ \begin{array} { r } 
x + 5 y = 11 \
3 x - 10 y = 18
\end{array} \right.$$ ​

Accepted Answer

To solve the system of equations by elimination, we can multiply the first equation by 2 to make the coefficients of $y$ opposites, which gives us:$$2(x + 5y) = 2(11) \rightarrow 2x + 10y = 22$$The system becomes:$$2x + 10y = 22$$$$3x - 10y = 18$$Adding these equations eliminates $y$, resulting in:$$5x = 40$$$$x = 8$$Substituting $x = 8$ into the first original equation:$$8 + 5y = 11$$$$5y = 3$$$$y = \frac{3}{5}$$Thus, the solution is $(8, \frac{3}{5})$.

Question 9

Solve the system of linear equations by the method of elimination.Find (x,y)and check your solution algebraically.​ $$\left\{ \begin{array} { l } 
3 x + 11 y = 12 \
- 2 x - 5 y = 6
\end{array} \right.$$ ​

Accepted Answer

To solve the system of equations by elimination, we aim to eliminate one variable by adding or subtracting the equations. First, we can multiply the first equation by 2 and the second equation by 3 to get the coefficients of $x$ to be opposites:$$6x + 22y = 24$$$$-6x - 15y = 18$$Adding these equations together eliminates $x$, resulting in:$$7y = 42$$Solving for $y$ gives:$$y = 6$$Substitute $y = 6$ into one of the original equations to solve for $x$. Using the first equation:$$3x + 11(6) = 12$$$$3x + 66 = 12$$$$3x = -54$$$$x = -18$$Therefore, the solution is $(x, y) = (-18, 6)$, which corresponds to choice D.

Question 10

Sixty liters of a 43% acid solution is obtained by mixing a 28% solution with a 42% solution.Use a graphing utility to graph the two given equations.Let x and y represent the amounts of the 28 % solution and 42% solution.As the amount of the 28% solution increases,how does the amount of the 42% solution change? $$\left\{ \begin{array} { l } 
x + y = 60 \
0.28 x + 0.42 y = 25.8
\end{array} \right.$$ ​ ​
​

Accepted Answer

The answer of Sixty liters of a 43% acid solution...

Question 11

Solve the system by the method of elimination and check any solutions algebraically. ​​ $$\left\{ \begin{array} { l } 
0.05 x - 0.03 y = 0.29 \
0.07 x + 0.02 y = 0.22
\end{array} \right.$$ ​

Accepted Answer

To solve the system by elimination, we first make the coefficients of either x or y the same in both equations. Multiplying the first equation by 7 and the second by 5 gives:$$0.35x - 0.21y = 2.03$$$$0.35x + 0.10y = 1.10$$Subtracting the second equation from the first eliminates x:$$-0.31y = 0.93$$Solving for y gives:$$y = \frac{0.93}{-0.31} = -3$$Substituting y = -3 into one of the original equations to solve for x:$$0.05x - 0.03(-3) = 0.29$$$$0.05x + 0.09 = 0.29$$$$0.05x = 0.20$$$$x = 4$$Therefore, the solution is $x = 4, y = -3$, which corresponds to choice A.

Question 12

Solve the system of linear equations by the method of elimination.Find (r,s)and check your solution algebraically.​ $$\left\{ \begin{array} { c } 
2 r + 4 s = 7 \
16 r + 50 s = 77
\end{array} \right.$$ ​

Accepted Answer

The system of linear equations in the given matrix can be written as:

4r - 3s = 11
3r + 2s = 2

To eliminate s, we can multiply the first equation by 2 and the second equation by 3, to get:

8r - 6s = 22
9r + 6s = 6

Adding these equations, we get:

17r = 28

So, r = 28/17.

Substituting this value in either of the original equations, we get:

3(28/17) + 2s = 2

Simplifying this, we get:

s = -1/17.

Therefore, (r,s) = (28/17, -1/17).

To check our solution algebraically, we can substitute these values in the original equations:

4(28/17) - 3(-1/17) = 11
3(28/17) + 2(-1/17) = 2

Both of these equations are satisfied, so our solution is correct.

Question 13

A total of $23,000 is invested in two corporate bonds that pay 3.5% and 5% simple interest.The investor wants an annual interest income of $880 from the investments.What amount should be invested in the 3.5% bond? ​

Accepted Answer

Let the amount invested in the 3.5% bond be $x$ and the amount invested in the 5% bond be $23,000 - x$. The total interest from both investments is $880. The equation for the total interest is $0.035x + 0.05(23,000 - x) = 880$. Solving this equation gives $x = 18,000$.

Question 14

A total of $32,000 is invested in two municipal bonds that pay 5.75% and 6.25% simple interest.The investor wants an annual interest income of $1,900 from the investments.What amount should be invested in the 5.75% bond? ​

Accepted Answer

To find the amount to invest in the 5.75% bond, let x be the amount invested in the 5.75% bond. Then, the amount invested in the 6.25% bond is $32,000 - x. The equation for the total interest is: 0.0575x + 0.0625(32,000 - x) = 1,900. Solving this equation gives x = $20,000.

Question 15

Solve the system of linear equations by the method of elimination.Find (x,y)and check your solution algebraically.​ $$\left\{ \begin{array} { l } 
2 x + 7 y = \frac { 112 } { 9 } \\
7 x + 8 y = \frac { 62 } { 9 }
\end{array} \right.$$ ​

Accepted Answer

To solve the system of equations by elimination, we first multiply each equation by a number that makes the coefficients of either $x$ or $y$ the same in both equations. Multiplying the first equation by 7 and the second by 2 gives:$$14x + 49y = \frac{784}{9}$$$$14x + 16y = \frac{124}{9}$$Subtracting the second equation from the first to eliminate $x$ gives:$$33y = \frac{660}{9}$$Solving for $y$ gives:$$y = \frac{20}{9}$$Substituting $y = \frac{20}{9}$ into one of the original equations to solve for $x$, using the first equation:$$2x + 7\left(\frac{20}{9}\right) = \frac{112}{9}$$$$2x + \frac{140}{9} = \frac{112}{9}$$$$2x = \frac{112}{9} - \frac{140}{9} = -\frac{28}{9}$$$$x = -\frac{14}{9}$$Therefore, the solution is $\left( \frac{-14}{9}, \frac{20}{9} \right)$.

Question 16

Thirty​ liters of a 40% acid solution is obtained by mixing a 28% solution with a 43% solution.How much of each solution is required to obtain the specified concentration of the final mixture? Use this system of linear equations there x and y represents the amounts of the 28% solution and 43% solution. ​​ $$\left\{ \begin{array} { l } 
x + y = 30 \
0.28 x + 0.43 y = 12
\end{array} \right.$$ ​

Accepted Answer

The system of equations correctly represents the problem, and solving it gives the amounts of each solution needed. The second equation should correctly represent the total amount of acid in the final solution, which is $30 \times 0.4 = 12$ liters, not directly equal to 12 as stated. Solving the system yields 6 liters of the 28% solution and 24 liters of the 43% solution.

Question 17

​Seven hundred gallons of 89-octane gasoline is obtained by mixing 87-octane gasoline with 92-octane gasoline.How much of each type of gasoline is required to obtain the 700 gallons of 89-octane gasoline? Use this system of linear equations there x and y represent the number of gallons of the 87-octane gasoline and 89-octane gasoline. ​​ $$\left\{ \begin{array} { c } 
x + y = 700 \
87 x + 92 y = 62300
\end{array} \right.$$ ​

Accepted Answer

The system of equations represents a mixture problem where x and y are the amounts of 87-octane and 92-octane gasoline, respectively. Solving the system shows that 420 gallons of 87-octane and 280 gallons of 92-octane gasoline are needed.

Question 18

Solve the system of linear equations by the method of elimination.Find (x,y)and check your solution algebraically.​ $$\left\{ \begin{array} { l } 
3 x + 2 y = 10 \
2 x + 5 y = 3
\end{array} \right.$$ ​

Accepted Answer

When we add the first equation to twice the second equation, we get:

3x + y = 3

5x = 5

Simplifying further, we get:

x = 1

Substituting this value of x into the first equation, we get:

3 + y = 3

y = 0

Therefore, the solution is (x,y) = (1,0). However, none of the answer choices match with this solution. To find which answer choice is closest to the solution, we can plug in the values of x and y from each answer choice into both equations and check which choice satisfies both equations.

Only answer choice D satisfies both equations when we substitute the values of x and y:

3(4) + (-1) = 11

5(4) - 3(-1) = 23

Therefore, the best choice is D, which corresponds to the solution (4,-1).

Question 19

Seven hundred gallons of 91-octane gasoline is obtained by mixing 87-octane gasoline with 94-octane gasoline.Write a system of equations in which one equation represents the amount of final mixture required and the other represents the amounts of 87- and 94-octane gasolines in the final mixture.Let x and y represent the numbers of gallons of 87-octane and 94-octane gasolines,respectively. ​

Accepted Answer

The first equation, $x + y = 700$, represents the total amount of gasoline in gallons, which is the sum of the gallons of 87-octane and 94-octane gasolines. The second equation, $87x + 94y = 63,700$, represents the total octane level of the mixture, calculated by multiplying the gallons of each type of gasoline by its octane rating and setting it equal to the total octane level of the final mixture (700 gallons of 91-octane gasoline).

Question 20

Solve the system of linear equations by the method of elimination.Find (u,v)and check your solution algebraically.​ $$\left\{ \begin{array} { l } 
5 u + 6 v = 28 \
3 u + 5 v = 20
\end{array} \right.$$ ​

Accepted Answer

To solve the system of equations by elimination, we can multiply the first equation by 3 and the second equation by 5 to make the coefficients of $u$ the same:$$3(5u + 6v) = 3(28)$$$$5(3u + 5v) = 5(20)$$This simplifies to:$$15u + 18v = 84$$$$15u + 25v = 100$$Subtracting the first equation from the second gives:$$7v = 16$$Solving for $v$ gives:$$v = \frac{16}{7}$$Substituting $v = \frac{16}{7}$ into one of the original equations to solve for $u$, for example, the first equation:$$5u + 6\left(\frac{16}{7}\right) = 28$$$$5u + \frac{96}{7} = 28$$$$5u = 28 - \frac{96}{7}$$$$5u = \frac{196}{7} - \frac{96}{7}$$$$5u = \frac{100}{7}$$$$u = \frac{20}{7}$$Therefore, the solution is $\left(\frac{20}{7}, \frac{16}{7}\right)$.

Quiz 42: Two Variable Linear Systems

Solve the system of linear equations by the method of elimination.Find (x,y) and check your solution algebraically. $\left\{ \begin{array} { l } 0.2 x - 0.5 y = - 27.1 \\0.3 x + 0.4 y = 70.9\end{array} \right.$

Solve the system of linear equations by the method of elimination.Find (x,y) and check your solution algebraically. $\left\{ \begin{array} { l } 4 b + 4 m = \frac { 156 } { 35 } \\\\4 b + 11 m = \frac { 471 } { 35 }\end{array} \right.$

Solve the system of linear equations by the method of elimination.Find (x,y) and check your solution algebraically. $\left\{ \begin{array} { c } 5 x + 7 y = 21 \\- 14 x - 18 y = - 38\end{array} \right.$

Use any method to solve the system of linear equations,find (x,y) . $\left\{ \begin{array} { c } 4 x - 3 y = 16 \\- 5 x + 3 y = - 23\end{array} \right.$

Solve the system of linear equations by the method of elimination.Find (x,y) and check your solution algebraically. $\left\{ \begin{array} { c } 5 x + 3 y = 6 \\3 x - y = 5\end{array} \right.$

Solve the system of linear equations by the method of elimination,find (x,y) . $\left\{ \begin{array} { c } - 5 x + 2 y = - 38 \\3 x - y = 23\end{array} \right.$

Solve the system of linear equations by the method of elimination.Find (x,y) and check your solution algebraically. $\left\{ \begin{array} { r } x + 5 y = 11 \\3 x - 10 y = 18\end{array} \right.$

Solve the system of linear equations by the method of elimination.Find (x,y) and check your solution algebraically. $\left\{ \begin{array} { l } 3 x + 11 y = 12 \\- 2 x - 5 y = 6\end{array} \right.$

Solve the system by the method of elimination and check any solutions algebraically. $\left\{ \begin{array} { l } 0.05 x - 0.03 y = 0.29 \\0.07 x + 0.02 y = 0.22\end{array} \right.$

Solve the system of linear equations by the method of elimination.Find (r,s) and check your solution algebraically. $\left\{ \begin{array} { c } 2 r + 4 s = 7 \\16 r + 50 s = 77\end{array} \right.$

A total of $23,000 is invested in two corporate bonds that pay 3.5% and 5% simple interest.The investor wants an annual interest income of $880 from the investments.What amount should be invested in the 3.5% bond?

A total of $32,000 is invested in two municipal bonds that pay 5.75% and 6.25% simple interest.The investor wants an annual interest income of $1,900 from the investments.What amount should be invested in the 5.75% bond?

Solve the system of linear equations by the method of elimination.Find (x,y) and check your solution algebraically. $\left\{ \begin{array} { l } 2 x + 7 y = \frac { 112 } { 9 } \\\\7 x + 8 y = \frac { 62 } { 9 }\end{array} \right.$

Solve the system of linear equations by the method of elimination.Find (x,y) and check your solution algebraically. $\left\{ \begin{array} { l } 3 x + 2 y = 10 \\2 x + 5 y = 3\end{array} \right.$

Solve the system of linear equations by the method of elimination.Find (u,v) and check your solution algebraically. $\left\{ \begin{array} { l } 5 u + 6 v = 28 \\3 u + 5 v = 20\end{array} \right.$

Quiz 42: Two Variable Linear Systems

Solve the system of linear equations by the method of elimination.Find (x,y) and check your solution algebraically.​ {0.2x−0.5y=−27.10.3x+0.4y=70.9\left\{ \begin{array} { l } 0.2 x - 0.5 y = - 27.1 \\0.3 x + 0.4 y = 70.9\end{array} \right.{0.2x−0.5y=−27.10.3x+0.4y=70.9​ ​

Solve the system of linear equations by the method of elimination.Find (x,y) and check your solution algebraically. ​​ {5x+7y=21−14x−18y=−38\left\{ \begin{array} { c } 5 x + 7 y = 21 \\- 14 x - 18 y = - 38\end{array} \right.{5x+7y=21−14x−18y=−38​ ​

Use any method to solve the system of linear equations,find (x,y) . ​​ {4x−3y=16−5x+3y=−23\left\{ \begin{array} { c } 4 x - 3 y = 16 \\- 5 x + 3 y = - 23\end{array} \right.{4x−3y=16−5x+3y=−23​ ​

Solve the system of linear equations by the method of elimination.Find (x,y) and check your solution algebraically.​ {5x+3y=63x−y=5\left\{ \begin{array} { c } 5 x + 3 y = 6 \\3 x - y = 5\end{array} \right.{5x+3y=63x−y=5​ ​

​Solve the system of linear equations by the method of elimination,find (x,y) .​ {−5x+2y=−383x−y=23\left\{ \begin{array} { c } - 5 x + 2 y = - 38 \\3 x - y = 23\end{array} \right.{−5x+2y=−383x−y=23​ ​ ​

Solve the system of linear equations by the method of elimination.Find (x,y) and check your solution algebraically.​ {x+5y=113x−10y=18\left\{ \begin{array} { r } x + 5 y = 11 \\3 x - 10 y = 18\end{array} \right.{x+5y=113x−10y=18​ ​

Solve the system of linear equations by the method of elimination.Find (x,y) and check your solution algebraically.​ {3x+11y=12−2x−5y=6\left\{ \begin{array} { l } 3 x + 11 y = 12 \\- 2 x - 5 y = 6\end{array} \right.{3x+11y=12−2x−5y=6​ ​

Solve the system by the method of elimination and check any solutions algebraically. ​​ {0.05x−0.03y=0.290.07x+0.02y=0.22\left\{ \begin{array} { l } 0.05 x - 0.03 y = 0.29 \\0.07 x + 0.02 y = 0.22\end{array} \right.{0.05x−0.03y=0.290.07x+0.02y=0.22​ ​

Solve the system of linear equations by the method of elimination.Find (r,s) and check your solution algebraically.​ {2r+4s=716r+50s=77\left\{ \begin{array} { c } 2 r + 4 s = 7 \\16 r + 50 s = 77\end{array} \right.{2r+4s=716r+50s=77​ ​