Answer:
The answer to your question is: (1/2, -1/3)
Step-by-step explanation:
Elimination
10x - 15y = 10 (I)
4x + 6y = 0 (II)
Multiply (I) by 2 and (II) by -5
20x - 30y = 20
-20x - 30y = 0
Add both equations
0 - 60y = 20
y = -20/60
y = - 1/3
Substitute y in (II)
4x + 6(-1/3) = 0
4x - 2 = 0
4x = 2
x = 2/4
x = 1/2
4(1/2) + 6(-1/3) = 0
4/2 - 6/3 = 0
2 - 2 = 0
0 = 0
Answer:
(1/2, -1/3)
Step-by-step explanation:
To solve using elimination, you have to make one of the variables from both equations additive inverses. Let's make the x-variables additive inverses.
First, we multiply the first equation by -2.
-2(10x - 15y) = -2(10)
-20x + 30y = -20
Now, we muliply the second equation by 5.
5(4x + 6y) = 0
20x + 30y = 0
If you look at the x-variables, now they are additive inverses. Now we add the two equations together.
-20x + 30y = -20
20x + 30y = 0
60y = -20
/60 /60
y = -1/3
Now you plug in the known y-value into any of the first equations. Let's use 4x + 6y = 0
4x + 6(-1/3) = 0
4x - 2 = 0
+2 +2
4x = 2
/4 /4
x = 1/2
The solution to this system of equations is (1/2, -1/3). You can change to decimal form if you want. To check your work, you plug in the x and y-values into the two equations in the question.
10x - 15y = 10
10(1/2) - 15(-1/3) = 10
5 + 5 = 10
10 = 10
So, the solution is true for this equation. Let's check the second one.
4x + 6y = 0
4(1/2) + 6(-1/3) = 0
2 - 2 = 0
0 = 0
Yes! When you plug the solution into the equations, they are true! So, your solution is (1/2, -1/3).
