Solve the system of two linear equations.

−3x−4y=−23
3x−5y=5 (1 point)

(5, 2)
left parenthesis 5 comma 2 right parenthesis

(5,−2)
left parenthesis 5 comma negative 2 right parenthesis

(−5, 2)
left parenthesis negative 5 comma 2 right parenthesis

(−5,−2)

(5,−2)

left parenthesis 5 comma negative 2 right parenthesis

To solve the system of two linear equations −3x−4y=−23 and 3x−5y=5, we can use the method of elimination or substitution. Let's use the method of elimination.

Step 1: Multiply the second equation by -1 to eliminate the x term.
-1(3x-5y) = -1(5)
-3x + 5y = -5

Step 2: Add the equations together to eliminate the x terms.
(-3x - 4y) + (-3x + 5y) = -23 + (-5)
-6x + y = -28 -----(Equation 3)

Step 3: Solve Equation 3 for y.
y = -28 + 6x

Step 4: Substitute the value of y in Equation 3 into either of the original equations.
-3x - 4(-28 + 6x) = -23
-3x + 112 - 24x = -23
-27x + 112 = -23
-27x = -23 - 112
-27x = -135
x = -135 / -27
x = 5

Step 5: Substitute the value of x into the expression for y obtained in Step 3.
y = -28 + 6(5)
y = -28 + 30
y = 2

Therefore, the solution to the given system of linear equations is (5, 2).

To solve the system of two linear equations, we can use the method of elimination or substitution. Let's solve it using the elimination method:

Step 1: Multiply Equation 1 by 3 and Equation 2 by -3 to eliminate x.
-9x - 12y = -69
-9x + 15y = -15

Step 2: Subtract Equation 1 from Equation 2 to eliminate x.
(3x - 5y) - (-3x - 4y) = 5 - (-23)
3x - 5y + 3x + 4y = 5 + 23
6x - y = 28

Now we have a new equation: 6x - y = 28.

Step 3: Solve the new equation for either x or y.
Let's solve it for y:
y = 6x - 28

Step 4: Substitute the value of y in terms of x back into one of the original equations.
-3x - 4(6x - 28) = -23
-3x - 24x + 112 = -23
-27x = -23 - 112
-27x = -135
x = -135 / -27
x = 5

Step 5: Substitute the value of x back into one of the original equations to solve for y.
-3(5) - 4y = -23
-15 - 4y = -23
-4y = -23 + 15
-4y = -8
y = -8 / -4
y = 2

Therefore, the solution to the system of equations is (5, 2).