Solve the system of two liner equations -3x-4y=-23 and 3x-5y=5

Answer 1

To solve the system of linear equations, we can use either the substitution method or the elimination method. I will use the elimination method.

First, let's multiply the second equation by 2 to eliminate x:
2*(3x - 5y) = 2*5
6x - 10y = 10

Now we have the system of equations:
-3x - 4y = -23 (Equation 1)
6x - 10y = 10 (Equation 2)

Next, add Equation 1 and Equation 2:
(-3x - 4y) + (6x - 10y) = (-23) + (10)
3x - 14y = -13 (Equation 3)

Now we have the system of equations:
-3x - 4y = -23 (Equation 1)
3x - 14y = -13 (Equation 3)

To eliminate x, multiply Equation 1 by 3 and Equation 3 by -3:
3*(-3x - 4y) = 3*(-23)
-9x - 12y = -69 (Equation 4)

-3*(3x - 14y) = -3*(-13)
-9x + 42y = 39 (Equation 5)

Now we have the system of equations:
-9x - 12y = -69 (Equation 4)
-9x + 42y = 39 (Equation 5)

Next, add Equation 4 and Equation 5:
(-9x - 12y) + (-9x + 42y) = (-69) + (39)
-18x + 30y = -30

Simplify the equation:
-6x + 10y = -10 (Equation 6)

So, now we have the system of equations:
-6x + 10y = -10 (Equation 6)
3x - 14y = -13 (Equation 3)

Now, let's multiply Equation 6 by -5:
-5*(-6x + 10y) = -5*(-10)
30x - 50y = 50 (Equation 7)

Now we have the system of equations:
30x - 50y = 50 (Equation 7)
3x - 14y = -13 (Equation 3)

Next, add Equation 7 and Equation 3:
(30x - 50y) + (3x - 14y) = 50 + (-13)
33x - 64y = 37

So now we have the system of equations:
33x - 64y = 37 (Equation 8)
3x - 14y = -13 (Equation 3)

Now, we can solve this system of equations. Subtract Equation 3 multiplied by 11 from Equation 8:
33x - 64y - (3x - 14y) * 11 = 37 - (-13) * 11
33x - 64y - 33x + 154y = 37 + 143
-64y + 154y = 37 + 143
90y = 180
y = 180/90
y = 2

Now substitute y = 2 into Equation 3 and solve for x:
3x - 14*2 = -13
3x - 28 = -13
3x = -13 + 28
3x = 15
x = 15/3
x = 5

Therefore, the solution to the system of equations -3x - 4y = -23 and 3x - 5y = 5 is x = 5 and y = 2.