To solve the system of equations, we can substitute the value of x from the first equation into the second equation.
x = 6y - 3
2(6y - 3) + 7y = 13
12y - 6 + 7y = 13
19y - 6 = 13
19y = 19
y = 1
Now that we have the value of y, we can substitute it back into the first equation to find x.
x = 6(1) - 3
x = 6 - 3
x = 3
Therefore, the solution to the system of equations is (3, 1).
Solve the system of equations. x=6y−3 2x+7y=13(1 point) (, )
7 answers
Solve the system of equations. 8x−3y= −22 y=10+4x (1 point) (, )
To solve this system of equations, we can substitute the value of y from the second equation into the first equation.
8x - 3(10 + 4x) = -22
8x - 30 - 12x = -22
-4x - 30 = -22
-4x = 8
x = -2
Now that we have the value of x, we can substitute it back into the second equation to find y.
y = 10 + 4(-2)
y = 10 - 8
y = 2
Therefore, the solution to the system of equations is (-2, 2).
8x - 3(10 + 4x) = -22
8x - 30 - 12x = -22
-4x - 30 = -22
-4x = 8
x = -2
Now that we have the value of x, we can substitute it back into the second equation to find y.
y = 10 + 4(-2)
y = 10 - 8
y = 2
Therefore, the solution to the system of equations is (-2, 2).
Solve the system of equations. 5x+2y=14 x−6y=22 (1 point) (, )
To solve the system of equations, we can use the method of substitution or elimination. Let's use the method of elimination in this case.
Given equations:
1) 5x + 2y = 14
2) x - 6y = 22
To eliminate one of the variables, we can multiply the second equation by 2 and add the equations together to eliminate y:
2(x - 6y) = 2(22)
2x - 12y = 44
Now, add this equation to the first equation:
5x + 2y + 2x - 12y = 14 + 44
7x - 10y = 58
Now we have one equation with only x and y. Next, we can solve for x:
7x - 10y = 58
7x = 10y + 58
x = (10/7)y + 58/7
Now, substitute this expression for x back into the first equation:
5((10/7)y + 58/7) + 2y = 14
(50/7)y + 290/7 + 2y = 98
(50/7)y + 2(7/7)y = 98 - 290/7
(64/7)y = (686 - 290) / 7
(64/7)y = 196 / 7
(64/7)y = 28
y = 28 * (7/64)
y = 3
Now, substitute y back into the equation to find x:
x - 6(3) = 22
x - 18 = 22
x = 22 + 18
x = 40
Therefore, the solution to the system of equations is (40, 3).
Given equations:
1) 5x + 2y = 14
2) x - 6y = 22
To eliminate one of the variables, we can multiply the second equation by 2 and add the equations together to eliminate y:
2(x - 6y) = 2(22)
2x - 12y = 44
Now, add this equation to the first equation:
5x + 2y + 2x - 12y = 14 + 44
7x - 10y = 58
Now we have one equation with only x and y. Next, we can solve for x:
7x - 10y = 58
7x = 10y + 58
x = (10/7)y + 58/7
Now, substitute this expression for x back into the first equation:
5((10/7)y + 58/7) + 2y = 14
(50/7)y + 290/7 + 2y = 98
(50/7)y + 2(7/7)y = 98 - 290/7
(64/7)y = (686 - 290) / 7
(64/7)y = 196 / 7
(64/7)y = 28
y = 28 * (7/64)
y = 3
Now, substitute y back into the equation to find x:
x - 6(3) = 22
x - 18 = 22
x = 22 + 18
x = 40
Therefore, the solution to the system of equations is (40, 3).
wrong
Let's try solving the system of equations again:
Given equations:
1) 5x + 2y = 14
2) x - 6y = 22
Let's solve by elimination:
Multiply the second equation by 2 to make the coefficients of x opposite in sign:
2(x - 6y) = 2(22)
2x - 12y = 44
Now, add the first equation to the modified second equation:
5x + 2y + 2x - 12y = 14 + 44
7x - 10y = 58
Now, isolate x in terms of y:
7x = 10y + 58
x = (10y + 58) / 7
Now, substitute this expression back into the first equation to solve for y:
5(10y + 58) / 7 + 2y = 14
50y + 290 + 14y = 98
64y + 290 = 98
64y = -192
y = -192 / 64
y = -3
Now, substitute y back into the second equation to solve for x:
x - 6(-3) = 22
x + 18 = 22
x = 22 - 18
x = 4
Therefore, the solution to the system of equations is (4, -3).
Given equations:
1) 5x + 2y = 14
2) x - 6y = 22
Let's solve by elimination:
Multiply the second equation by 2 to make the coefficients of x opposite in sign:
2(x - 6y) = 2(22)
2x - 12y = 44
Now, add the first equation to the modified second equation:
5x + 2y + 2x - 12y = 14 + 44
7x - 10y = 58
Now, isolate x in terms of y:
7x = 10y + 58
x = (10y + 58) / 7
Now, substitute this expression back into the first equation to solve for y:
5(10y + 58) / 7 + 2y = 14
50y + 290 + 14y = 98
64y + 290 = 98
64y = -192
y = -192 / 64
y = -3
Now, substitute y back into the second equation to solve for x:
x - 6(-3) = 22
x + 18 = 22
x = 22 - 18
x = 4
Therefore, the solution to the system of equations is (4, -3).