solve 3u_5v=_11 and 2u+7v=3 using elimination method
I'm assuming that _ means -, since they are on the same key, so you have
3u-5v = -11
2u+7v = 3
To eliminate y, do
21u-35v = -77
10u+35v = 15
and add to get
31u = -62
Now finish it off
3u-5v = -11
2u+7v = 3
Multiply Eq1 by 2 and Eq2 by 3; then subtract Eq2 from Eq1:
6u-10v = -22
6u+21v = 9
Diff: -31v = -31
V = 1.
In Eq1, replace v with 1; and solve for u.
To solve the system of equations using the elimination method, we need to eliminate one variable by manipulating the equations. Here are the steps to solve the given system:
Equation 1: 3u - 5v = -11
Equation 2: 2u + 7v = 3
Step 1: Multiply Equation 1 by 7 and Equation 2 by 5 to make the coefficients of 'v' in both equations the same:
7(3u - 5v) = 7(-11) --> 21u - 35v = -77 (Multiplying Equation 1 by 7)
5(2u + 7v) = 5(3) --> 10u + 35v = 15 (Multiplying Equation 2 by 5)
Step 2: Add the equations together to eliminate the 'v' variable:
(21u - 35v) + (10u + 35v) = -77 + 15
31u = -62
Step 3: Divide both sides of the equation by 31 to solve for 'u':
31u/31 = -62/31
u = -2
Step 4: Substitute the value of 'u' (-2) into either of the original equations to solve for 'v'. We'll use Equation 2:
2(-2) + 7v = 3
-4 + 7v = 3
7v = 3 + 4
7v = 7
v = 1
Therefore, the solution to the system of equations is u = -2 and v = 1.
To solve the system of equations using the elimination method, we need to eliminate either "u" or "v" from the equations. Let's begin:
1) Multiply the first equation (3u - 5v = -11) by 2:
2 * (3u - 5v) = 2 * (-11)
6u - 10v = -22
2) Now, let's write down the second equation (2u + 7v = 3).
3) Add the modified equation (step 1) to the second equation (step 2):
(6u - 10v) + (2u + 7v) = -22 + 3
6u - 10v + 2u + 7v = -19
4) Combine like terms:
8u - 3v = -19
Now, we have a new equation: 8u - 3v = -19.
To find the value of "u" or "v" from this equation, we need another equation. Let's use the original second equation (2u + 7v = 3).
5) Multiply the second equation (2u + 7v = 3) by 5:
5 * (2u + 7v) = 5 * 3
10u + 35v = 15
6) Now, let's write down the modified equation (10u + 35v = 15).
7) Subtract the modified equation (step 6) from the new equation (step 4):
(8u - 3v) - (10u + 35v) = -19 - 15
8u - 10u - 3v - 35v = -34
8) Combine like terms:
-2u - 38v = -34
Now, we have another new equation: -2u - 38v = -34.
We have a system of equations consisting of two new equations:
1) 8u - 3v = -19
2) -2u - 38v = -34
Now, we can solve this new system of equations using any method (substitution, elimination, or matrix).
Let's use the elimination method again to eliminate "u":
1) Multiply the second equation (-2u - 38v = -34) by 4:
4 * (-2u - 38v) = 4 * (-34)
-8u - 152v = -136
2) Now, let's write down the first equation (8u - 3v = -19).
3) Add the modified equation (step 1) to the first equation (step 2):
(8u - 3v) + (-8u - 152v) = -19 + (-136)
8u - 3v - 8u - 152v = -155
4) Combine like terms:
-155v = -155
5) Solve for "v":
v = -155 / -155
v = 1
Now that we have the value of "v," we can substitute it back into one of the original equations to find the value of "u." Let's use the second equation (2u + 7v = 3):
2u + 7(1) = 3
2u + 7 = 3
2u = 3 - 7
2u = -4
u = -4 / 2
u = -2
Therefore, the solution to the system of equations is u = -2 and v = 1.