Solve the system of equations.
4x−8y=−36
−6x+y=32
To solve this system of equations, we can use the method of substitution.
First, we can rearrange the second equation to solve for y:
-6x + y = 32
y = 6x + 32
Next, we can substitute this expression for y into the first equation:
4x - 8y = -36
4x - 8(6x + 32) = -36
Simplifying this equation, we get:
4x - 48x - 256 = -36
-44x = 220
x = -5
Now that we know the value of x, we can substitute it back into either of the original equations to solve for y. Using the second equation:
-6x + y = 32
-6(-5) + y = 32
y = 2
Therefore, the solution to the system of equations is:
x = -5
y = 2
6x−3y=−6
9x−3y=−18
We can use the method of elimination to solve this system of equations.
First, we'll write the equations in standard format:
6x - 3y = -6
9x - 3y = -18
Notice that both equations have a common term of -3y. To eliminate this term, we can subtract the first equation from the second:
(9x - 3y) - (6x - 3y) = -18 - (-6)
3x = -12
x = -4
Now that we know the value of x, we can substitute it back into either equation to solve for y. Using the first equation:
6x - 3y = -6
6(-4) - 3y = -6
-24 - 3y = -6
y = 6
Therefore, the solution to the system of equations is:
x = -4
y = 6
5x–4y=–44
–x+3y=22
We can use the method of substitution to solve this system of equations:
From the second equation, we can solve for x in terms of y:
-x + 3y = 22
x = 3y - 22
Now, we can substitute this expression for x into the first equation:
5x - 4y = -44
5(3y - 22) - 4y = -44
15y - 110 - 4y = -44
11y = 66
y = 6
We've now found the value of y, so we can substitute this back into either equation to solve for x. Using the second equation:
-x + 3y = 22
-x + 3(6) = 22
-x + 18 = 22
x = -4
Therefore, the solution to the system of equations is:
x = -4
y = 6
−6x+2y=−32
4x+3y=−9
We can use the method of elimination to solve this system of equations:
First, we'll multiply the first equation by -2 to eliminate the y term:
-2(-6x + 2y = -32)
12x - 4y = 64
Now we can add this equation to the second equation to eliminate the x term:
12x - 4y = 64
4x + 3y = -9
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16x - y = 55
Now we can solve for y by subtracting 16x from both sides:
16x - y = 55
-y = 55 - 16x
y = -55 + 16x
We can substitute this expression for y back into either of the original equations to solve for x. Using the first equation:
-6x + 2y = -32
-6x + 2(-55 + 16x) = -32
-6x - 110 + 32x = -32
26x = 78
x = 3
Now we can substitute x = 3 into our expression for y:
y = -55 + 16x
y = -55 + 16(3)
y = -7
Therefore, the solution to the system of equations is:
x = 3
y = -7
The Math Club sold slices of pizza, p, at a profit of $3 each and cans of soda, s, at $2 each, to raise money for a trip. They sold 200 items and made a profit of $550. Write the pair of linear equations that model this situation.(1 point)
Equation 1, per item: = 200
Equation 2, total: = 550
Let's use the following variables:
p = number of pizza slices sold
s = number of cans of soda sold
From the problem, we know the following:
- The Math Club sold 200 items in total:
p + s = 200
- The profit from selling each slice of pizza is $3, and they sold p slices:
3p
- The profit from selling each can of soda is $2, and they sold s cans:
2s
- The total profit made was $550:
3p + 2s = 550
Therefore, the pair of linear equations that model this situation are:
p + s = 200
3p + 2s = 550