Solve the system by elimination. State whether the system has one solution, infinite solutions, or no solution.
x + 3y = –5
4x – y = 6
please help thanks
eliminate one of the variables by adjusting coefficients so you can cancel (eliminate)
multiplying the 2nd equation by 3, gives the y-term a -3 coefficient
... 12x - 3y = 18
adding the 1st equation eliminates the y-terms
find x , then substitute back to find y
Sure, I can help! Let's solve the system by elimination.
First, let's multiply the first equation by 4 to eliminate the x-term in the second equation:
4(x + 3y) = 4(-5)
4x + 12y = -20
Now we have the system:
4x + 12y = -20
4x - y = 6
To eliminate the x-term, we can subtract the second equation from the first:
4x + 12y - (4x - y) = -20 - 6
4x + 12y - 4x + y = -26
Simplifying, we get:
13y = -26
Dividing both sides by 13, we find:
y = -2
Now we can substitute this value of y into one of the original equations. Let's use the first equation:
x + 3(-2) = -5
x - 6 = -5
x = 1
So the solution to the system is x = 1 and y = -2. Therefore, the system has one solution.
And there you have it! If you have any more questions, feel free to ask, and I'll be happy to help!
To solve the system of equations by elimination, we need to manipulate the equations in a way that allows us to add or subtract them to eliminate one variable.
Let's start by multiplying the first equation by 4 to get rid of the x coefficient in the second equation.
4(x + 3y) = 4(-5)
Simplifying, we get:
4x + 12y = -20
Now, we can subtract the second equation (4x - y = 6) from this equation (4x + 12y = -20):
(4x + 12y) - (4x - y) = -20 - 6
Simplifying further:
4x + 12y - 4x + y = -26
13y = -26
Now, we can solve for y by dividing both sides of the equation by 13:
y = -26 / 13
y = -2
Now that we know the value of y, we can substitute it back into one of the original equations to solve for x. Let's use the first equation:
x + 3(-2) = -5
x - 6 = -5
x = -5 + 6
x = 1
So the solution to the system of equations is x = 1 and y = -2.
Therefore, the system has one solution.
To solve the system by elimination, we need to eliminate one variable by multiplying one or both equations by appropriate constants so that the coefficients of one of the variables become additive inverses.
Let's eliminate the variable y by multiplying the first equation by 4 and the second equation by 3:
4(x + 3y) = 4(-5)
3(4x - y) = 3(6)
This simplifies to:
4x + 12y = -20
12x - 3y = 18
Now, we can add the two equations together to eliminate the y variable:
(4x + 12y) + (12x - 3y) = -20 + 18
Simplifying this equation gives:
16x + 9y = -2
Now, we have a new equation that only involves the variables x and y.
To determine whether the system has one solution, infinite solutions, or no solution, we need to compare the number of variables to the number of equations. In this case, we have two variables (x and y) and three equations (the original two equations and the new equation).
Since the number of equations is greater than the number of variables, we can potentially have a unique solution, infinite solutions, or no solution. To determine which case it is, we would need to solve the new equation and check if it is consistent with the original system.