4x-3y=5
Solve {
5y+7=3x
using the method of elimination.
rearrange things a bit to get
x-3y=5
3x-5y=7
If you multiply the 1st equation by 3, you have
3x-9y=15
3x-5y=7
Now just subtract to eliminate x and you get
-4y = 8
y = -2
Now use either equation to solve for x.
Had that first equation not mysteriously changed we would have;
1st times 3 ---> 12x - 9y = 15
2nd times 4 ---> 12 - 20y = 28
subtract:
11y = -13
y = -13/11
into 1st:
4x -3(-13/11) = 5
4x = 5 - 39/11 = 16/11
x = 4/11
To solve the system of equations using the method of elimination, we need to eliminate one variable by either adding or subtracting the equations. Let's start by manipulating the equations in a way that will allow us to cancel out one of the variables.
Given system of equations:
Equation 1: 4x - 3y = 5
Equation 2: 5y + 7 = 3x
We can start by multiplying Equation 1 by 3 and Equation 2 by 4 to make the coefficients of x in both equations the same, which will help in eliminating the variable x.
3 * (Equation 1): 12x - 9y = 15
4 * (Equation 2): 20y + 28 = 12x
Now, we can subtract Equation 2 (20y + 28 = 12x) from Equation 1 (12x - 9y = 15) to eliminate x:
(12x - 9y) - (20y + 28) = 15 - 28
12x - 9y - 20y - 28 = -13
12x - 29y - 28 = -13
Simplifying the equation, we get:
12x - 29y = 15 -- (Equation 3)
Now, we have Equation 3 (12x - 29y = 15) and Equation 2 (5y + 7 = 3x) to work with. We can eliminate the variable x by multiplying Equation 2 by 4 and Equation 3 by 3.
4 * (Equation 2): 20y + 28 = 12x
3 * (Equation 3): 36x - 87y = 45
Now, we can subtract Equation 2 (20y + 28 = 12x) from Equation 3 (36x - 87y = 45) to eliminate x:
(36x - 87y) - (20y + 28) = 45 - 28
36x - 87y - 20y - 28 = 17
36x - 107y - 28 = 17
Simplifying the equation, we get:
36x - 107y = 45 -- (Equation 4)
Now, we have Equation 3 (12x - 29y = 15) and Equation 4 (36x - 107y = 45) to work with. We can eliminate the variable x by multiplying Equation 3 by 3 and Equation 4 by 4.
3 * (Equation 3): 36x - 87y = 45
4 * (Equation 4): 144x - 428y = 180
Now, we can subtract Equation 3 (36x - 87y = 45) from Equation 4 (144x - 428y = 180) to eliminate x:
(144x - 428y) - (36x - 87y) = 180 - 45
144x - 428y - 36x + 87y = 135
108x - 515y = 135 -- (Equation 5)
Now, we are left with Equation 5 (108x - 515y = 135) which only has x and y as variables. We can now solve for one of the variables and substitute back to find the other.
Let's solve Equation 5 for x:
108x = 135 + 515y
x = (135 + 515y) / 108 -- (Equation 6)
Now, we can substitute Equation 6 back into either Equation 3 or Equation 4 to solve for y.
Let's substitute Equation 6 into Equation 3 (12x - 29y = 15):
12 * ((135 + 515y) / 108) - 29y = 15
(1620 + 2060y) / 108 - 29y = 15
(1620 + 2060y - 3132y) / 108 = 15
(1620 - 1072y) / 108 = 15
To solve for y, cross multiply:
1620 - 1072y = 15 * 108
1620 - 1072y = 1620
-1072y = 0
y = 0
Now, substitute the value of y = 0 back into Equation 6 to solve for x:
x = (135 + 515(0)) / 108
x = 135 / 108
x = 1.25
Therefore, the solution to the system of equations is x = 1.25 and y = 0.