To solve this system of equations, we can use either the substitution method or the elimination method.
Here, we will use the elimination method to solve the system of equations:
1. Multiply the first equation by 4 and the second equation by 3 to make the coefficient of y the same in both equations:
4(3x + y = 2) becomes 12x + 4y = 8
3(7x - 4y = 30) becomes 21x - 12y = 90
2. Add the two equations together to eliminate y:
12x + 4y + 21x - 12y = 8 + 90
33x - 8y = 98
3. Solve for x:
33x - 8y = 98
33x - (8)(3x) = 98
33x - 24x = 98
9x = 98
x = 98 / 9
x = 10.888
4. Substitute the value of x back into one of the original equations to solve for y. We will use the first equation:
3x + y = 2
3(10.888) + y = 2
32.664 + y = 2
y = 2 - 32.664
y = -30.664
Therefore, the solution to the system of equations is x = 10.888 and y = -30.664.
3X+y=2
7x-4y=30
Solve the system of equations
7 answers
No decimals
To solve the system of equations without decimals, we can use the elimination method:
1. Multiply the first equation by 4 and the second equation by 3 to make the coefficient of y the same in both equations:
4(3x + y = 2) becomes 12x + 4y = 8
3(7x - 4y = 30) becomes 21x - 12y = 90
2. Add the two equations together to eliminate y:
12x + 4y + 21x - 12y = 8 + 90
33x - 8y = 98
3. Solve for x:
33x - 8y = 98
33x - 8(3x) = 98
33x - 24x = 98
9x = 98
x = 98 / 9
x = 10
4. Substitute the value of x back into one of the original equations to solve for y. We will use the first equation:
3x + y = 2
3(10) + y = 2
30 + y = 2
y = 2 - 30
y = -28
Therefore, the solution to the system of equations is x = 10 and y = -28.
1. Multiply the first equation by 4 and the second equation by 3 to make the coefficient of y the same in both equations:
4(3x + y = 2) becomes 12x + 4y = 8
3(7x - 4y = 30) becomes 21x - 12y = 90
2. Add the two equations together to eliminate y:
12x + 4y + 21x - 12y = 8 + 90
33x - 8y = 98
3. Solve for x:
33x - 8y = 98
33x - 8(3x) = 98
33x - 24x = 98
9x = 98
x = 98 / 9
x = 10
4. Substitute the value of x back into one of the original equations to solve for y. We will use the first equation:
3x + y = 2
3(10) + y = 2
30 + y = 2
y = 2 - 30
y = -28
Therefore, the solution to the system of equations is x = 10 and y = -28.
It’s wrong.. write it like this (__,__)
Let's solve the system again using the elimination method:
1. The given equations are:
3x + y = 2 ...(1)
7x - 4y = 30 ...(2)
2. Multiply equation (1) by 4 and equation (2) by 1 to eliminate y:
12x + 4y = 8 ...(3)
7x - 4y = 30 ...(4)
3. Add equations (3) and (4) to eliminate y:
19x = 38
x = 38 / 19
x = 2
4. Substitute the value of x back into equation (1) to find y:
3(2) + y = 2
6 + y = 2
y = 2 - 6
y = -4
Therefore, the solution to the system of equations is (2, -4).
1. The given equations are:
3x + y = 2 ...(1)
7x - 4y = 30 ...(2)
2. Multiply equation (1) by 4 and equation (2) by 1 to eliminate y:
12x + 4y = 8 ...(3)
7x - 4y = 30 ...(4)
3. Add equations (3) and (4) to eliminate y:
19x = 38
x = 38 / 19
x = 2
4. Substitute the value of x back into equation (1) to find y:
3(2) + y = 2
6 + y = 2
y = 2 - 6
y = -4
Therefore, the solution to the system of equations is (2, -4).
Solve the system of equations
2x+6y=-18
x=5y-1
2x+6y=-18
x=5y-1
To solve the system of equations, we can substitute the expression for x from the second equation into the first equation and solve for y.
Given equations:
1. 2x + 6y = -18
2. x = 5y - 1
Substitute x from equation 2 into equation 1:
2(5y - 1) + 6y = -18
10y - 2 + 6y = -18
16y - 2 = -18
16y = -16
y = -1
Now, substitute y back into equation 2 to find x:
x = 5(-1) - 1
x = -5 - 1
x = -6
Therefore, the solution to the system of equations is x = -6 and y = -1.
Given equations:
1. 2x + 6y = -18
2. x = 5y - 1
Substitute x from equation 2 into equation 1:
2(5y - 1) + 6y = -18
10y - 2 + 6y = -18
16y - 2 = -18
16y = -16
y = -1
Now, substitute y back into equation 2 to find x:
x = 5(-1) - 1
x = -5 - 1
x = -6
Therefore, the solution to the system of equations is x = -6 and y = -1.
2 answers