Solve the system of equations. x=6y−3 2x+7y=13
To solve the system of equations, we can use the substitution method or the elimination method.
Let's solve using the substitution method first:
From the first equation, we have x = 6y - 3.
We can substitute this expression for x in the second equation and solve for y.
Substituting x into the second equation, we get:
2(6y - 3) + 7y = 13
12y - 6 + 7y = 13
19y - 6 = 13
19y = 13 + 6
19y = 19
y = 1
Now that we have the value of y, we can substitute it back into the first equation to solve for 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 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 can subtract this new equation from the first equation to eliminate x.
(5x + 2y) - (5x - 30y) = 14 - 110
5x + 2y - 5x + 30y = -96
32y = -96
y = -3
Now that we have the value of y, we can substitute it back into either equation to solve for x. Let's substitute it into the first equation.
5x + 2(-3) = 14
5x - 6 = 14
5x = 14 + 6
5x = 20
x = 4
Therefore, the solution to the system of equations is x = 4 and y = -3.
To solve the system of equations, we can again use the elimination method.
First, let's multiply the first equation by 4 and the second equation by -1 to make the coefficients of y in both equations the same but with opposite signs.
4(-9x + y) = 4(12)
-36x + 4y = 48
-1(2x + 4y) = -1(-28)
-2x - 4y = 28
Now, we can add the two equations together to eliminate y.
(-36x + 4y) + (-2x - 4y) = 48 + 28
-36x + 4y - 2x - 4y = 76
-38x = 76
x = -2
Now that we have the value of x, we can substitute it back into either equation to solve for y. Let's substitute it into the first equation.
-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 the number of solutions for a system of linear equations, we can check the determinant of the coefficient matrix.
The coefficient matrix for this system is:
\[
\begin{bmatrix}
6 & 2 \\
3 & 1 \\
\end{bmatrix}
\]
The determinant of this matrix is calculated as follows: (6 * 1) - (2 * 3) = 6 - 6 = 0.
Since the determinant is equal to zero, this indicates that there is no unique solution for the system. The system of equations is either inconsistent or dependent, meaning there are either infinitely many solutions or no solutions at all.
Solve the system of equations. 8x−3y= −22 y=10+4x
We can solve the system of equations by substitution.
From the second equation, we have y = 10 + 4x.
We can substitute this expression for y in the first equation and solve for x.
Substituting y into the first equation, we get:
8x - 3(10 + 4x) = -22
8x - 30 - 12x = -22
-4x - 30 = -22
-4x = -22 + 30
-4x = 8
x = -2
Now that we have the value of x, we can substitute it back into the second equation to solve for y.
y = 10 + 4(-2)
y = 10 - 8
y = 2
Therefore, the solution to the system of equations is x = -2 and y = 2.
Solve the system of equations. 5x+2y=14 x−6y=22
Solve the system of equations. –9x+y=12 2x+4y= −28
How many solutions does the system of equations have? 6x+2y=−18 3x+y=24
Solve the system of equations. 2x+6y=−18 x=5y−1
To solve the system of equations, we can use the substitution method.
Let's substitute the expression for x in the first equation:
2(5y - 1) + 6y = -18
10y - 2 + 6y = -18
16y - 2 = -18
16y = -18 + 2
16y = -16
y = -1
Now that we know the value of y, we can substitute it back into the second equation to solve for x:
x = 5(-1) - 1
x = -5 - 1
x = -6
Therefore, the solution to the system of equations is x = -6 and y = -1.
Solve the system of equations. 3x+y=2 7x−4y=30
To solve the system of equations, we can use the elimination method.
First, let's multiply the first equation by 4 to make the coefficients of y in both equations the same but with opposite signs.
4(3x + y) = 4(2)
12x + 4y = 8
Now, we can add the two equations together to eliminate y.
(12x + 4y) + (7x - 4y) = 8 + 30
12x + 7x + 4y - 4y = 38
19x = 38
x = 38/19
x = 2
Now that we have the value of x, we can substitute it back into the first equation to solve for y.
3(2) + y = 2
6 + y = 2
y = 2 - 6
y = -4
Therefore, the solution to the system of equations is x = 2 and y = -4.
Solve the system of equations. −4x+6y=38 x−5y=−20
To solve the system of equations, we can use the elimination method.
First, let's multiply the second equation by 4 to make the coefficients of x in both equations opposites.
4(x - 5y) = 4(-20)
4x - 20y = -80
Now, let's add the two equations together to eliminate x.
(-4x + 6y) + (4x - 20y) = 38 + (-80)
-4x + 4x + 6y - 20y = -42
-14y = -42
y = -42 / -14
y = 3
Now that we have the value of y, we can substitute it back into either equation to solve for x. Let's substitute it into the second equation.
x - 5(3) = -20
x - 15 = -20
x = -20 + 15
x = -5
Therefore, the solution to the system of equations is x = -5 and y = 3.
How many solution(s) does the system of equations have? −5x+y=−18 10x−2y=36
To determine the number of solutions for a system of linear equations, we can check the determinant of the coefficient matrix.
The coefficient matrix for this system is:
\[
\begin{bmatrix}
-5 & 1 \\
10 & -2 \\
\end{bmatrix}
\]
The determinant of this matrix is calculated as follows: (-5 * -2) - (1 * 10) = 10 - 10 = 0.
Since the determinant is equal to zero, this indicates that there is no unique solution for the system. The system of equations is either inconsistent or dependent, meaning there are either infinitely many solutions or no solutions at all.
i only need one answer
Apologies for the confusion. Let us solve the system of equations.
We have the equations:
1) -5x + y = -18
2) 10x - 2y = 36
To eliminate y, we can multiply the first equation by 2 and add it to the second equation:
2 * (-5x + y) + (10x - 2y) = -18*2 + 36
-10x + 2y + 10x - 2y = -36 + 36
0 = 0
Since 0 = 0, the two equations are dependent, meaning there are infinitely many solutions. Every ordered pair (x, y) that satisfies the first equation will also satisfy the second equation.
Hence, there are infinitely many solutions to the system of equations.