Review the graphs of a system of two linear equations in two variables: 4x+3y=15 and y=x+5 . Find the solution to both equations. (1 point)
To find the solution to both equations, we can solve the system of equations by substitution or elimination.
First, let's solve the system by substitution:
From the second equation, we can express y in terms of x:
y = x + 5
Now, substitute this expression for y in the first equation:
4x + 3(x + 5) = 15
4x + 3x + 15 = 15
7x + 15 = 15
7x = 0
x = 0
Now, substitute this value of x back into the second equation to find the corresponding y-value:
y = 0 + 5
y = 5
So the solution to the system of equations is x = 0, y = 5.
Now, let's review the graphs of the two equations:
The first equation, 4x + 3y = 15, can be written in slope-intercept form as y = (-4/3)x + 5. This equation represents a line with a slope of -4/3 and a y-intercept of 5. Since the coefficient of x is negative, the line slopes downward from left to right.
The second equation, y = x + 5, is already in slope-intercept form. This equation represents a line with a slope of 1 and a y-intercept of 5.
To graph these equations, plot several points and connect them with a straight line. Alternatively, you can easily identify the y-intercept of each equation and use the slope to find additional points on the line.
The graphs of these two equations intersect at the point (0, 5), which confirms our solution to the system of equations.
Review the graphs of a system of two linear equations in two variables: y=x−6 and y=−x+4 . Find the solution to both equations. (1 point)
To find the solution to both equations, we can solve the system of equations by substitution or elimination.
First, let's solve the system by substitution:
From the first equation, we can express y in terms of x:
y = x - 6
Now, substitute this expression for y in the second equation:
x - 6 = -x + 4
2x = 10
x = 5
Now, substitute this value of x back into the first equation to find the corresponding y-value:
y = 5 - 6
y = -1
So the solution to the system of equations is x = 5, y = -1.
Now, let's review the graphs of the two equations:
The first equation, y = x - 6, is in slope-intercept form. This equation represents a line with a slope of 1 and a y-intercept of -6. This line slopes upward from left to right.
The second equation, y = -x + 4, is also in slope-intercept form. This equation represents a line with a slope of -1 and a y-intercept of 4. This line slopes downward from left to right.
To graph these equations, plot several points and connect them with a straight line. Alternatively, you can easily identify the y-intercept of each equation and use the slope to find additional points on the line.
The graphs of these two equations intersect at the point (5, -1), which confirms our solution to the system of equations.
Estimate the intersection point of the graphs of the following linear equations in two variables.
y=x+2
y=−x+6
To find the intersection point of the two graphs, we need to set the two equations equal to each other and solve for x:
x + 2 = -x + 6.
Adding x to both sides of the equation:
2x + 2 = 6.
Subtracting 2 from both sides:
2x = 4.
Dividing both sides by 2:
x = 2.
Now that we have the value of x, we substitute it into either equation to find the y-coordinate of the intersection point. Let's use the first equation, y = x + 2:
y = 2 + 2 = 4.
Therefore, the intersection point of the two graphs is (2, 4).
Estimate the intersection point of the graphs of the following linear equations in two variables.
y=−x+7
y=4x−3
To find the intersection point of the two graphs, we need to set the two equations equal to each other and solve for x:
-x + 7 = 4x - 3.
Adding x to both sides of the equation:
7 = 5x - 3.
Adding 3 to both sides:
10 = 5x.
Dividing both sides by 5:
x = 2.
Now that we have the value of x, we substitute it into either equation to find the y-coordinate of the intersection point. Let's use the first equation, y = -x + 7:
y = -2 + 7 = 5.
Therefore, the intersection point of the two graphs is (2, 5).
Find the coordinates of the intersection point that solves the system of these two linear equations in two variables: 2x+y=15 and y=−x+5 . (1 point) Responses (5,−10) left parenthesis 5 comma negative 10 right parenthesis (−5, 10) left parenthesis negative 5 comma 10 right parenthesis (10,−5) left parenthesis 10 comma negative 5 right parenthesis (−10, 5)
To find the intersection point of the two equations, we need to set them equal to each other and solve for x and y.
First, let's substitute the value of y from the second equation into the first equation:
2x + (-x + 5) = 15
Simplifying the equation:
2x - x + 5 = 15
x + 5 = 15
Subtracting 5 from both sides:
x = 10
Now, substitute this value of x back into either equation to find the corresponding y-value. Let's use the second equation, y = -x + 5:
y = -(10) + 5
y = -10 + 5
y = -5
Therefore, the coordinates of the intersection point that solves the system of equations are (10, -5).