determine the number of solutions for the system of linear equations
y = -10x + 5
300 = -15x + 60y
To find the number of solutions for the system of linear equations, we need to determine if they intersect at a single point, are parallel lines, or are coincident lines.
First, let's rewrite the equations in standard form:
10x + y = 5 (equation 1)
15x - 60y = -300 (equation 2)
To find the number of solutions, we can compare the slopes of the lines represented by the equations.
The slope of equation 1 is -10, and the slope of equation 2 is 1/4 (15/60).
Since the slopes are different, the lines represented by the equations are not parallel.
Now, let's solve the system of equations to see if they intersect at a single point:
10x + y = 5 (equation 1)
15x - 60y = -300 (equation 2)
Rearranging equation 1, we have:
y = -10x + 5
Substituting this into equation 2, we get:
15x - 60(-10x + 5) = -300
15x + 600x - 3000 = -300
615x - 3000 = -300
615x = 2700
x = 2700/615
x = 4.39 (approx)
Substituting x = 4.39 back into equation 1:
y = -10(4.39) + 5
y = -43.9 + 5
y = -38.9
Therefore, the system of linear equations intersects at a single point (4.39, -38.9).
The number of solutions for the system of linear equations is 1.
solve = -8x + 32 = -16
To solve the equation -8x + 32 = -16, we can follow these steps:
1. Start with the given equation: -8x + 32 = -16
2. Subtract 32 from both sides to isolate the term with x:
-8x + 32 - 32 = -16 - 32
-8x = -48
3. Divide both sides by -8 to solve for x:
-8x / -8 = -48 / -8
x = 6
Therefore, the solution to the equation -8x + 32 = -16 is x = 6.
To determine the number of solutions for the system of linear equations, we can use several methods such as graphing, substitution, or elimination. Let's use the elimination method to solve this system.
Given system of linear equations:
Equation 1: y = -10x + 5
Equation 2: 300 = -15x + 60y
To eliminate one variable, we need to multiply both sides of Equation 1 by 60 to make the coefficients of y equal:
Multiplying Equation 1 by 60:
60y = -600x + 300
Now we can rewrite the system with the modified Equation 1:
Equation 1: 60y = -600x + 300
Equation 2: 300 = -15x + 60y
Next, subtract the modified Equation 2 from Equation 1 to eliminate the variable y:
(60y - 60y) = (-600x + 300) - (-15x + 60y)
0 = -600x + 15x + 300 - 0
0 = -585x + 300
Simplifying the equation:
585x = 300
Divide both sides by 585:
x = 300 / 585
x ≈ 0.5128
Now substitute the value of x into one of the original equations (Equation 1) to find the value of y:
y = -10x + 5
y = -10 * 0.5128 + 5
y ≈ -4.128 + 5
y ≈ 0.872
Therefore, the system of linear equations has a single solution: x ≈ 0.5128, y ≈ 0.872.
To determine the number of solutions for a system of linear equations, we can use the method of substitution or elimination.
Let's begin by using the substitution method to solve the system of equations:
1) Start with the first equation: y = -10x + 5.
2) Substitute this expression into the second equation for y:
300 = -15x + 60(-10x + 5).
3) Simplify the equation:
300 = -15x + 60(-10x) + 60(5).
300 = -15x - 600x + 300.
4) Combine like terms:
300 = -615x + 300.
5) Subtract 300 from both sides of the equation:
0 = -615x.
6) Divide both sides of the equation by -615:
x = 0.
Now that we have found the value of x, we can substitute it back into one of the original equations to find the value of y:
Using the first equation, y = -10x + 5:
y = -10(0) + 5,
y = 0 + 5,
y = 5.
Therefore, we have found that the solution to the system of linear equations is x = 0 and y = 5.
Since we have found a unique solution (x = 0, y = 5), there is only one solution to the system of linear equations.