Solve the system of two linear equations.
−3x−4y=−23
3x−5y=5 (1 point)
Responses
(−5,−2)
left parenthesis negative 5 comma negative 2 right parenthesis
(−5, 2)
left parenthesis negative 5 comma 2 right parenthesis
(5,−2)
left parenthesis 5 comma negative 2 right parenthesis
(5, 2)
___________________________________
2.) How many solutions does the system of equations have?
8x−6y=48
8x+6y=48 (1 point)
Responses
two
two
infinite
infinite
one
one
none
___________________________________3.) Solve the system of two equations using elimination with multiplication.
4x−6y=−48
−2x+4y=30 (1 point)
Responses
(−33,−9)
left parenthesis negative 33 comma negative 9 right parenthesis
(−21,−6)
left parenthesis negative 21 comma negative 6 right parenthesis
(−3, 6)
left parenthesis negative 3 comma 6 right parenthesis
(3, 9)
___________________________________4.)Solve the system of equations.
3x−4y=40
5x−5y=55 (1 point)
Responses
(18, 7)
left parenthesis 18 comma 7 right parenthesis
(14, 3)
left parenthesis 14 comma 3 right parenthesis
(4,−7)
left parenthesis 4 comma negative 7 right parenthesis
(8,−3)
___________________________________5.) How many solutions does the system have?
12x−15y=18
4x−5y=6 (1 point)
Responses
two solutions
two solutions
one solution
one solution
no solution
no solution
infinitely many solutions
1.) To solve the system of two linear equations:
−3x−4y=−23
3x−5y=5
Add the two equations together:
−3x − 4y + 3x − 5y = −23 + 5
−9y = −18
Divide by -9 to get:
y = 2
Now, substitute y = 2 into one of the original equations to find x:
−3x − 4(2) = −23
−3x − 8 = −23
−3x = −15
x = 5
The solution is (5, 2).
2.) The system of equations has:
8x−6y=48
8x+6y=48
These equations are parallel and represent the same line since if you divide the second equation by 2, you get the first equation. Therefore, there are infinitely many solutions.
3.) For the elimination with multiplication, first align the coefficients of x or y by multiplying the necessary equations:
4x−6y=−48
−2x+4y=30
Multiply the second equation by 2:
4x−6y=−48
−4x+8y=60
Now, add the two equations together:
(4x−6y) + (−4x+8y) = (−48) + (60)
2y = 12
y = 6
Substitute y = 6 back into one of the original equations to find x:
4x−6(6)=−48
4x−36=−48
4x=−12
x=−3
The solution is (−3, 6).
4.) To solve the system of equations:
3x−4y=40
5x−5y=55
We can use either substitution or elimination. For convenience, let's multiply the first equation by 5 and the second equation by 3:
15x−20y=200
15x−15y=165
Now, subtract the second equation from the first:
(15x−20y) − (15x−15y) = (200) − (165)
−5y = 35
y = −7
Substitute y = −7 back into one of the original equations to find x:
3x−4(−7)=40
3x+28=40
3x=12
x=4
The solution is (4,−7).
5.) When we have:
12x−15y=18
4x−5y=6
By inspection, the second equation is just 1/3 of the first equation. This means these two equations represent the same line. Therefore, the system has infinitely many solutions.