Solve this system of linear equaiton by an algebraic method.
So If I had to solve this the elimination method way, how would I do it?
x-2y = 10
3x-y = 0
x - 2y = 10
3x - y = 0
eliminate y by multiplying 3x - y = 0 by -2
-2(3x - y = 0) = -6x + 2y = 0
ADD the 2 equations
x - 2y = 10
-6x + 2y = 0
-5x + 0 = 10
-5x = 10
x = -2
substitute x = -2 in x - 2y = 10, to find y
x - 2y = 10
-2 - 2y = 10
-2y = 12
y = -6
check x = -2, y = -6
3x - y = 0
3(-2) - -6 = 0
-6 + 6 = 0
0 = 0
Oh so You have to multiply it by -2 because the first equation also uses -2 and you have to cancel them out that way right?
So if its - and + that means you have to add the equations. If its - and - that means you have to subtract and if it's + and + that also means you have to subtract right?
How did you get -2 as an answer?
Did you divide 5x by 5 and 10 by five to get 2?
You lost me right here:
How did you do this?
x - 2y = 10
-2 - 2y = 10
-2y = 12
y = -6
How did you get the second line?
-2 - 2y = 10?
And at the end you went 0 = 0
so they have to equal to zero always?
How did you get -2 as an answer?
Did you divide 5x by 5 and 10 by five to get 2? YES
You lost me right here:
How did you do this?
x - 2y = 10
-2 - 2y = 10
-2y = 12
y = -6
How did you get the second line?
-2 - 2y = 10?
to find the value of y, since you know x = -2, you plug x = -2 into either equation and solve for y
the second line is from plugging -2 in for x in the equation x - 2y = 10
x = -2
x - 2y = 10
-2 - 2y = 10
-2y = 12
y = -6
yes, when you check, both sides will equal 0 if you are correct
To solve this system of linear equations using the elimination method, you can follow these steps:
Step 1: Multiply one or both of the equations by a suitable number(s) so that the coefficients of one of the variables (x or y) in both equations become equal but with opposite signs. Let's eliminate the y term first.
Multiply equation 1 by 3 and equation 2 by 2 to make the y coefficients equal and opposite in sign:
(3) * (x - 2y) = (3) * 10 --> 3x - 6y = 30
(2) * (3x - y) = (2) * 0 --> 6x - 2y = 0
Now we have the equations:
3x - 6y = 30
6x - 2y = 0
Step 2: Add or subtract the two equations to eliminate one variable. In this case, subtracting the second equation from the first will cancel out the y term:
(3x - 6y) - (6x - 2y) = 30 - 0
3x - 6y - 6x + 2y = 30
-3x - 4y = 30
Simplifying gives:
-3x - 4y = 30
Step 3: Solve the resulting equation for the remaining variable. In this case, let's solve for x:
-3x = 30 + 4y
x = (30 + 4y) / -3
x = (-30/3) - (4y/3)
x = -10 - (4/3)y
Step 4: Substitute the value of x obtained in Step 3 into one of the original equations (either equation 1 or 2). Let's substitute it into equation 1:
x - 2y = 10
(-10 - (4/3)y) - 2y = 10
-10 - (4/3)y - 2y = 10
-10 - (10/3)y = 10
Step 5: Solve the resulting equation for y:
-10 - (10/3)y = 10
(10/3)y = -10 - 10
(10/3)y = -20
y = (-20 * 3) / 10
y = -6
Step 6: Substitute the value of y obtained in Step 5 into one of the original equations. Let's use equation 1:
x - 2y = 10
x - 2(-6) = 10
x + 12 = 10
x = 10 - 12
x = -2
Therefore, the solution to the system of equations is x = -2 and y = -6.