Question

Solve by substitution. Choose all that apply.

2x + y = 5
3x - 3y = 3

Select one or more:
A. The solution has a negative y value.
B. The solution has both a negative x and a negative y.
C. The solution has a negative x value.
D. The solution has a positive x value.
E. no solution

Solve by substitution.

2x - y = 7

6x - 3y = 14

Select one:
A. No solution
B. The solution has a negative x value.
C. Infinitely many solutions
D. The solution has a positive x value.
Solve by elimination.

2x - y = 7

3x + y = 8

Select one:
A. The solution has two positive values.
B. The solution has a negative x value.
C. The solution has one positive and one negative value.
D. The solution has a positive y value.

Solve by elimination.

3x - y = 12

5x + 2y = 20

Select one:
A. The solution has one value that is zero.
B. The solution has two positive values.
C. The solution has one positive and one negative value.
D. The solution has two negative values.

Solve by any method you choose.

3x + y = 8

3x + 3/2 y = 12

Select one:
a. There are no solutions to the system.
b. There is one solution to the system.
c. There are infinitely many solutions to the system.

Answer 1

Answer:

1. Correct choice is D, x=2, y=1 (the solution has a positive x value).

2. Correct choice is A,  (no solution)

3. Correct choice is C, x=3, y=-1 (one positive, one negative value)

4. Correct choice is A, x=4, y=0 (the solution has one value that is zero.)

5. Correct choice is B, x=0, y=8 (there is one solution to the system)

Step-by-step explanation:

1. Solve by substitution.

$\left\{\begin{array}{l}2x + y = 5\\3x - 3y = 3\end{array}\right.\Rightarrow \left\{\begin{array}{l}2(y+1) + y = 5\\x=y +1\end{array}\right.\Rightarrow \left\{\begin{array}{l}2y+2+y=5\\x=y+1\end{array}\right.\Rightarrow \\ \\\left\{\begin{array}{l}3y=3\\x=y+1\end{array}\right.\Rigtharrow \left\{\begin{array}{l}y=1\\x=2\end{array}\right..$

2. Solve by substitution.

$\left\{\begin{array}{l}2x- y = 7\\6x - 3y = 14\end{array}\right.\Rightarrow \left\{\begin{array}{l}y=2x-7\\6x-3(2x-7)=14\end{array}\right.\Rightarrow \left\{\begin{array}{l}y=2x-7\\6x-6x+21=14\end{array}\right.\Rightarrow \\ \\\left\{\begin{array}{l}y=2x-7\\21=14\end{array}\right..$

Since 21 never equal to 14, then the system has no solution.

3. Solve by elimination.

$\left\{\begin{array}{l}2x-y=7\\3x+y=8\end{array}\right..$

Add both equations:

$2x-y+3x+y=7+8,\\ \\5x=15,\\ \\x=3.$

Multiply first equation by 3 and second equation by 2 and subtract them:

$6x-3y-(6x+2y)=21-16,\\ \\6x-3y-6x-2y=5,\\ \\-5y=5,\\ \\y=-1.$

4. Solve by elimination.

$\left\{\begin{array}{l}3x-y=12\\5x+2y=20\end{array}\right..$

Multiply first equation by 2 and add both equations:

$6x-2y+5x+2y=24+20,\\ \\11x=44,\\ \\x=4.$

Multiply first equation by 5 and second equation by 3 and subtract them:

$15x-5y-(15x+6y)=60-60,\\ \\15x-5y-15x-6y=0,\\ \\-11y=0,\\ \\y=0.$

5. Solve by elimination and substitution.

$\left\{\begin{array}{l}3x+y=8\\ \\3x+\dfrac{3}{2}y=12\end{array}\right..$

Subtract from the first equation the second one:

$3x+y-\left(3x+\dfrac{3}{2}\right)=8-12,\\ \\3x+y-3x-\dfrac{3}{2}y=-4,\\ \\-\dfrac{1}{2}y=-4,\\ \\y=8.$

Substitute y=8 into the first equation:

$3x+8=8,\\ \\3x=0,\\ \\x=0.$