Solve by the linear combination method (with or without multiplication).
4x-5y =-18
3x + 2y =-2
Eq1: 4X - 5Y = -18
Eq2: 3X + 2Y = -2
Multiply Eq1 by -3 and Eq2 by 4:
-12X + 15Y = 54
12X + 8Y = -8
Add the 2 Eqs and get:
23Y = 46,
Y = 46 / 23 = 2.
Substitute 2 for Y in Eq2:
3X + 2*2 = -2,
3X + 4 = -2,
3X = -6,
X = -2.
Solution Set = (-2 , 2).
To solve this system of equations using the linear combination method, we'll aim to eliminate one variable by multiplying one or both equations. Then, we can solve for the remaining variable and substitute that value back into one of the original equations to find the other variable.
Let's start by multiplying the second equation by 2 to make the coefficients of y in both equations equal.
2(3x + 2y) = 2(-2)
6x + 4y = -4
Now, we have the system of equations:
4x - 5y = -18
6x + 4y = -4
To eliminate y, we'll add the two equations together:
(4x - 5y) + (6x + 4y) = -18 + (-4)
Simplifying the equation:
10x - y = -22
Now, we have a new equation with only one variable, x.
Next, we can solve for x by isolating it:
10x = -22 + y
x = (-22 + y)/10
Now that we have x in terms of y, we can substitute this expression into one of the original equations. Let's use the first equation:
4x - 5y = -18
Substituting x:
4((-22 + y)/10) - 5y = -18
(-88 + 4y)/10 - 5y = -18
Multiplying throughout by 10 to eliminate the denominator:
-88 + 4y - 50y = -180
-88 - 46y = -180
Adding 88 to both sides:
-46y = -92
Dividing by -46:
y = 2
Now that we have the value of y, we can substitute it back into the expression we found for x:
x = (-22 + y)/10
x = (-22 + 2)/10
x = -2/10
x = -0.2
Therefore, the solution to the system of equations is x = -0.2 and y = 2.