Use linear combinations to solve the system of linear equations.
5x=4y-30
2x+3y+-12
5x = 4y - 30
2x + 3y +-12
(1) 5x - 4y = -30
(2) 2x + 3y = 12
multiply (1) by 3(5x - 4y = -30)
(1) 15x + 12y = -90
multiply (2) by 4(2x + 3y = 12)
(2) 8x + 12y = 48
ADD two new equations
(1) 15x + 12y = -90
(2) 8x + 12y = 48
23x + 0 = -42
23x = -42
x = -42/23
substitute x = -42/23 in 2x + 3y = 12
to find y
(2) 2x + 3y = 12
2(-42/23) + 3y = 12
-84/23 + 3y = 12
3y = 12 + 84/23
3y = 276/23 + 84/23
3y = 360/23
y = 360/23 * 1/3
y = 360/69
y = 120/23
x = -42/23,
y = 120/23
To solve the system of linear equations using linear combinations, we need to eliminate one variable by multiplying one or both equations by appropriate constants such that the coefficients of one variable in both equations will cancel each other out.
Let's start with the given equations:
1) 5x = 4y - 30
2) 2x + 3y = -12
We can see that the coefficients of y in both equations are currently different, so let's eliminate the y variable by multiplying equation (2) by 4 and equation (1) by 3.
Multiplying equation (2) by 4:
4(2x + 3y) = 4(-12)
8x + 12y = -48 (new equation)
Multiplying equation (1) by 3:
3(5x) = 3(4y - 30)
15x = 12y - 90
15x - 12y = -90 (new equation)
Now we have two new equations:
3) 8x + 12y = -48
4) 15x - 12y = -90
To eliminate the y variable, we'll add equations (3) and (4) together:
(8x + 12y) + (15x - 12y) = (-48) + (-90)
8x + 15x + 12y - 12y = -48 - 90
23x = -138
Now we have a new equation:
5) 23x = -138
Next, we can solve equation (5) for x:
23x = -138
x = -138/23
x = -6
Now that we have the value of x, we can substitute it back into one of the original equations to find the value of y. Let's choose equation (1):
5x = 4y - 30
5(-6) = 4y - 30
-30 = 4y - 30
4y = 0
Divide both sides of the equation by 4:
4y/4 = 0/4
y = 0
Therefore, the solution to the system of linear equations is x = -6 and y = 0.
To solve the system of linear equations using linear combinations, we need to eliminate one variable by multiplying one or both equations by a constant so that the coefficients of that variable are the same with opposite signs. Then, we can add or subtract the equations to eliminate that variable.
Let's start with the given equations:
1) 5x = 4y - 30
2) 2x + 3y = -12
We'll start by eliminating the x variable.
To do this, we need to ensure that the coefficients of x are the same but with opposite signs in both equations. Currently, the coefficient of x in equation 1 is 5, and there is no term with x in equation 2. So, we need to eliminate the x term from equation 2.
To make the coefficient of x in equation 2 match equation 1, we can multiply equation 2 by 5:
5 × (2x + 3y) = 5 × (-12) -> 10x + 15y = -60
Now we have two equations:
1) 5x = 4y - 30
2) 10x + 15y = -60
Next, we'll multiply equation 1 by 2 to make the coefficients of x match equation 2:
2 × (5x) = 2 × (4y - 30) -> 10x = 8y - 60
Now we have:
1) 10x = 8y - 60
2) 10x + 15y = -60
We can subtract equation 1 from equation 2 to eliminate x:
(10x + 15y) - (10x) = (-60) - (8y - 60)
15y = -60 - 8y + 60
15y + 8y = 0
23y = 0
Solving for y, we get y = 0.
Now we substitute the value of y = 0 into equation 1:
10x = 8(0) - 60
10x = -60
x = -6
So the solution to the system of linear equations is x = -6 and y = 0.