i don't get these please help me:
use the subtitution method to solve the linear system
1)3x+y=0 x-y=4
2)y=-3x-1 x-3y=3
3)3x-y=9 2x+y=6
4)x-2y=0 3x+y=0
Thank you!!:)
1)3x + y = 0, x - y = 4
Pick and equation, and solve for either x or y
x - y = 4
x = 4 + y
Substitute, x = 4 + y, for x, in the other equation and solve for y
x = 4 + y
3x + y = 0
3(4 + y) + y = 0
12 + 3y + y = 0
12 + 4y = 0
4y = -12
y = -3
substitute y = -3 in either equation to find x
x - y = 4
x - (-3) = 4
x + 3 = 4
x = 1
Now you try the others
so are these correct?
2)y=-3x-1 x-3y=3
x=0 y=-1???
3)3x-y=9 2x+y=6
dont know help??
4)x-2y=0 3x+y=0
please help!!! i don't know how you changed x - y = 4 to x = 4 + y tell me please
Thanks!!
2) you are correct
x - y = 4
add + y to both sides
x - y + y = 4 + y
x = 4 + y
Sure! I'll explain how to use the substitution method to solve each of the given linear systems.
To use the substitution method, follow these steps:
1) Solve one of the equations for one variable in terms of the other variable.
2) Substitute the expression found in step 1 into the other equation.
3) Solve the resulting equation for the remaining variable.
4) Substitute the value obtained in step 3 back into either of the original equations to find the value of the other variable.
5) Express the solution as an ordered pair (x, y).
Let's apply the substitution method to each of the given linear systems:
1) 3x + y = 0
x - y = 4
From the second equation, we can solve for x in terms of y:
x = y + 4
Substitute this expression for x in the first equation:
3(y + 4) + y = 0
Solve for y:
3y + 12 + y = 0
4y + 12 = 0
4y = -12
y = -3
Substitute y = -3 back into the second equation to solve for x:
x - (-3) = 4
x + 3 = 4
x = 1
Therefore, the solution to the first linear system is (1, -3).
2) y = -3x - 1
x - 3y = 3
Since the first equation is already solved for y, we can substitute it into the second equation:
x - 3(-3x - 1) = 3
Simplify the equation:
x + 9x + 3 = 3
10x + 3 = 3
10x = 0
x = 0
Substitute x = 0 back into the first equation to solve for y:
y = -3(0) - 1
y = -1
Thus, the solution to the second linear system is (0, -1).
3) 3x - y = 9
2x + y = 6
Solve the second equation for y:
y = 6 - 2x
Substitute this expression for y in the first equation:
3x - (6 - 2x) = 9
Simplify the equation:
3x - 6 + 2x = 9
5x - 6 = 9
5x = 15
x = 3
Substitute x = 3 back into the second equation to solve for y:
2(3) + y = 6
6 + y = 6
y = 0
Hence, the solution to the third linear system is (3, 0).
4) x - 2y = 0
3x + y = 0
Solve the first equation for x:
x = 2y
Substitute this expression for x in the second equation:
3(2y) + y = 0
Simplify the equation:
6y + y = 0
7y = 0
y = 0
Substitute y = 0 back into the first equation to solve for x:
x - 2(0) = 0
x = 0
Therefore, the solution to the fourth linear system is (0, 0).
I hope this helps! Let me know if you have any further questions.