Solve the system of two linear equations. −3x−4y=−23 3x−5y=5 (1 point) Responses (5, 2) left parenthesis 5 comma 2 right parenthesis (−5, 2) left parenthesis negative 5 comma 2 right parenthesis (−5,−2) left parenthesis negative 5 comma negative 2 right parenthesis (5,−2) left parenthesis 5 comma negative 2 right parenthesis
To solve the system of equations, we can use the method of elimination. We'll eliminate the variable "x" by adding the two equations.
(-3x - 4y) + (3x - 5y) = -23 + 5
-3x + 3x - 4y - 5y = -18
-9y = -18
Now, let's solve for y by dividing both sides of the equation by -9.
-9y / -9 = -18 / -9
y = 2
Next, substitute the value of y back into either of the original equations. Let's use the second equation:
3x - 5(2) = 5
3x - 10 = 5
3x = 15
Now, solve for x by dividing both sides of the equation by 3.
3x / 3 = 15 / 3
x = 5
Therefore, the solution to the system of equations is (5, 2).
To solve the system of equations −3x−4y=−23 and 3x−5y=5, we can use the method of substitution.
First, let's solve one equation for one variable in terms of the other variable. Let's solve the second equation for x:
3x − 5y = 5
We can rearrange this equation to isolate x:
3x = 5 + 5y
Divide both sides by 3:
x = (5 + 5y) / 3
Now, substitute the value of x into the first equation:
-3((5 + 5y) / 3) - 4y = -23
Multiply both sides by 3 to get rid of the fractions:
-3(5 + 5y) - 12y = -69
Distribute the -3:
-15 - 15y - 12y = -69
Combine like terms:
-27y = -69 + 15
Simplify:
-27y = -54
Divide both sides by -27:
y = -54 / -27
Simplify:
y = 2
Now that we know y = 2, we can substitute this value back into the equation we solved for x:
x = (5 + 5y) / 3
x = (5 + 5(2)) / 3
x = (5 + 10) / 3
x = 15 / 3
Simplify:
x = 5
Therefore, the solution to the system of equations is (5, 2).
To solve the system of two linear equations −3x−4y=−23 and 3x−5y=5, we can use the method of substitution or elimination. I will explain how to use the elimination method.
1. Start with the two given equations:
−3x−4y=−23 ...(Equation 1)
3x−5y=5 ...(Equation 2)
2. Multiply Equation 1 by 3 and Equation 2 by -3 to eliminate the x term:
−9x−12y=−69 ...(Equation 3) (Multiply Equation 1 by 3)
−9x+15y=−15 ...(Equation 4) (Multiply Equation 2 by -3)
3. Add Equation 3 and Equation 4 together to eliminate the x term:
−9x−12y+(-9x+15y)=−69+(-15)
-9x - 9x - 12y + 15y = -69 - 15
-18x + 3y = -84 ...(Equation 5)
4. Now we have a new equation (Equation 5): -18x + 3y = -84 that only involves y.
5. Solve Equation 5 for y:
-18x + 3y = -84
3y = -18x - 84
y = (-18x - 84)/3
y = -6x - 28 ...(Equation 6)
6. Substitute Equation 6 into any of the original equations to solve for x. Let's use Equation 1:
−3x−4(-6x - 28) = −23
−3x+24x+112 = −23
21x + 112 = -23
21x = -23 - 112
21x = -135
x = -135/21
x = -45/7
7. Now we have the value of x. Substitute this value into Equation 6 to find the value of y:
y = -6x - 28
y = -6(-45/7) - 28
y = 270/7 - 196/7
y = 74/7
8. Therefore, the solution to the system of equations −3x−4y=−23 and 3x−5y=5 is (x, y) = (-45/7, 74/7), which can be written as (-45/7, 10.6) rounded to one decimal place.
So, the correct response is: (−5, 2) left parenthesis negative 5 comma 2 right parenthesis.