Solve the system of equations.
x = 6y - 3
2x + 7y = 13 (1 point)
To solve the system of equations, we can use the substitution method.
From the first equation, we have x = 6y - 3. We can substitute this expression for x in the second equation.
2(6y - 3) + 7y = 13
12y - 6 + 7y = 13
19y - 6 = 13
Add 6 to both sides:
19y = 19
Divide both sides by 19:
y = 1
Now, substitute this value of y 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 x = 3 and y = 1.
To solve the system of equations, we can use the substitution method.
From the second equation, we have y = 10 + 4x. We can substitute this expression for y in the first equation.
8x - 3(10 + 4x) = -22
8x - 30 - 12x = -22
Combine like terms:
-4x - 30 = -22
Add 30 to both sides:
-4x = 8
Divide both sides by -4:
x = -2
Now, substitute this value of x 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 x = -2 and y = 2.
To solve the system of equations, we can use the elimination method.
First, let's multiply the second equation by 5 to make the coefficients of x in both equations the same:
5(x - 6y) = 5(22)
5x - 30y = 110
Now, we have the following system of equations:
5x + 2y = 14 (Equation 1)
5x - 30y = 110 (Equation 2)
By subtracting Equation 1 from Equation 2, we can eliminate the x variable:
(5x - 30y) - (5x + 2y) = 110 - 14
5x - 30y - 5x - 2y = 96
-32y = 96
Divide both sides by -32:
y = 96 / -32
y = -3
Now substitute this value of y in Equation 1:
5x + 2(-3) = 14
5x - 6 = 14
5x = 14 + 6
5x = 20
Divide both sides by 5:
x = 20 / 5
x = 4
Therefore, the solution to the system of equations is x = 4 and y = -3.
To solve the system of equations, we can use the elimination method.
Multiply the first equation by 4 and the second equation by 1 so that the coefficients of y in both equations are the same:
4(-9x + y) = 4(12)
2x + 4y = -28
Now we have the following system of equations:
-36x + 4y = 48 (Equation 1)
2x + 4y = -28 (Equation 2)
By subtracting Equation 2 from Equation 1, we can eliminate the y variable:
(-36x + 4y) - (2x + 4y) = 48 - (-28)
-36x + 4y - 2x - 4y = 48 + 28
-38x = 76
Divide both sides by -38:
x = 76 / -38
x = -2
Now substitute this value of x in Equation 1:
-9(-2) + y = 12
18 + y = 12
y = 12 - 18
y = -6
Therefore, the solution to the system of equations is x = -2 and y = -6.
To determine how many solutions the system of equations has, we can rewrite the second equation as y = 24 - 3x (by subtracting 3x from both sides).
Now, we can compare the coefficients of x and y in both equations:
Equation 1: 6x + 2y = -18
Equation 2: 0x + 1y = 24 - 3x
We can see that the coefficients of x and y in the two equations are not equal, therefore the lines represented by the equations are not parallel.
Since the lines are not parallel, they must intersect at a single point and the system has exactly one solution.
Therefore, the system of equations has one unique solution.
Solve the system of equations. (1 point)
8x - 3y = - 22.
y = 10 + 4x (1 point))
Solve the system of equations.
5x + 2y = 14
x - 6y = 22 (1 point)
Solve the system of equations.
- 9x + y = 12
2x + 4y = - 28 (1 point)
How many solutions does the system of equations have?
6x + 2y = - 18
3x + y = 24 (1 point)
To solve the system of equations, we can use the method of substitution. Here's how you can do it:
Step 1: Solve one equation for one variable in terms of the other variable.
From the first equation, we can rearrange it to solve for x in terms of y:
x = 6y - 3
Step 2: Substitute the expression found in Step 1 into the other equation.
Substitute the expression for x in the second equation:
2(6y - 3) + 7y = 13
Step 3: Simplify and solve for y.
Distribute the 2:
12y - 6 + 7y = 13
Combine like terms:
19y - 6 = 13
Add 6 to both sides:
19y = 19
Divide by 19:
y = 1
Step 4: Use the value of y to find the value of x.
Substitute the value of y = 1 into the first equation:
x = 6(1) - 3
x = 6 - 3
x = 3
So the solution to the system of equations is x = 3 and y = 1.