Solving Systems of Equations by Graphing Practice
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Question
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A coordinate plane with four quadrants shows the x- and y-axes ranging from negative 5 to 5 in increments of 1. A solid line and a dotted line intersect each other. The equation of the solid line is x minus 5 y equals 3. The equation of the dotted line is 3 x minus 2 y equals negative 4. The intersection of both lines is shown at negative 2 on the x-axis and negative 1 on the y-axis in quadrant 3.
Review the graphs of a system of two linear equations in two variables: x−5y=7 and 3x−2y=−4. Find the solution to both equations.
(1 point)
The intersection point is (
).
My apologies for the mistake. Let's solve the system of equations correctly.
To find the solution to both equations, we need to find the coordinates of the intersection point.
First, let's rearrange the equations in slope-intercept form (y = mx + b) or isolate y.
For the equation x - 5y = 3:
-5y = -x + 3
Dividing by -5:
y = (1/5)x - 3/5
For the equation 3x - 2y = -4:
-2y = -3x - 4
Dividing by -2:
y = (3/2)x + 2
Now we have both equations in slope-intercept form. To find the intersection point, we need to equate the y values:
(1/5)x - 3/5 = (3/2)x + 2
Let's solve for x:
Multiply both sides by 10 to eliminate fractions:
2x - 6 = 15x + 20
Move the x terms to one side:
2x - 15x = 20 + 6
-13x = 26
Divide by -13:
x = -2
Now substitute x = -2 into one of the original equations, such as x - 5y = 3, to find y:
-2 - 5y = 3
-5y = 3 + 2
-5y = 5
y = -1
Therefore, the intersection point is (-2, -1).
To find the solution to the system of equations, we can solve them simultaneously.
First, let's rearrange the equations in slope-intercept form (y = mx + b) or isolate y.
For the equation 4x + 3y = 15:
3y = -4x + 15
y = (-4/3)x + 5
For the equation y = x + 5, it is already in slope-intercept form.
Now we can set the two equations equal to each other to solve for the intersection point:
(-4/3)x + 5 = x + 5
Let's solve for x:
Multiply both sides by 3 to eliminate the fraction:
-4x + 15 = 3x + 15
Move the x terms to one side:
-4x - 3x = 15 - 15
-7x = 0
Divide by -7:
x = 0
Now substitute x = 0 into one of the original equations, such as y = x + 5, to find y:
y = 0 + 5
y = 5
Therefore, the solution to the system of equations is (0, 5).
To find the solution to the system of equations, we can solve them simultaneously.
First, let's rearrange the equations in slope-intercept form (y = mx + b) or isolate y.
For the equation y = x - 6, it is already in slope-intercept form.
For the equation y = -x + 4, it is also already in slope-intercept form.
Now we can set the two equations equal to each other to solve for the intersection point:
x - 6 = -x + 4
Let's solve for x:
Add x to both sides:
2x - 6 = 4
Add 6 to both sides:
2x = 10
Divide by 2:
x = 5
Now substitute x = 5 into one of the original equations, such as y = x - 6, to find y:
y = 5 - 6
y = -1
Therefore, the solution to the system of equations is (5, -1).
To estimate the intersection point of the graphs of the given linear equations, we can set them equal to each other:
y = x + 2
y = -x + 6
Now we can solve for x:
x + 2 = -x + 6
Combine like terms:
2x = 4
Divide both sides by 2:
x = 2
Now substitute x = 2 into one of the original equations, such as y = x + 2, to find y:
y = 2 + 2
y = 4
Therefore, the estimated intersection point of the graphs is (2, 4).
To estimate the intersection point of the graphs of the given linear equations, we can set them equal to each other:
y = -x + 7
y = 4x - 3
Now we can solve for x:
-x + 7 = 4x - 3
Combine like terms:
5x = 10
Divide both sides by 5:
x = 2
Now substitute x = 2 into one of the original equations, such as y = -x + 7, to find y:
y = -(2) + 7
y = 5
Therefore, the estimated intersection point of the graphs is (2, 5).