Review the graph of a system of two linear equations in two variables X minus 5Y equals seven and 3X minus 2Y equals -4. Find a solution to both equations what is the intersection point
To review the graph of the system of equations, let's start by solving the system.
Given the system of equations:
1) X - 5Y = 7
2) 3X - 2Y = -4
To find the solution to both equations (the intersection point), we can use any method of solving systems of equations, such as substitution or elimination. In this case, let's use the method of elimination.
Multiply equation 1 by 3 and equation 2 by 5 to make the coefficients of X cancel each other when the equations are added together:
3(X - 5Y) = 3(7) -> 3X - 15Y = 21
5(3X - 2Y) = 5(-4) -> 15X - 10Y = -20
Now, add the equations together to eliminate X:
(3X - 15Y) + (15X - 10Y) = 21 + (-20)
18X - 25Y = 1
Now, we have a new equation:
3X - 2Y = -4 -> 18X - 12Y = -24
Next, we can solve the new equation for X:
18X - 25Y = 1 -> 18X = 1 + 25Y -> X = (1 + 25Y)/18
Substitute this value of X back into equation 1:
X - 5Y = 7 -> ((1 + 25Y)/18) - 5Y = 7
Multiply through by 18 to eliminate the fraction:
1 + 25Y - 90Y = 126 -> -65Y = 125
Solve for Y:
Y = 125/(-65) -> Y = -25/13
Substitute this value of Y back into the expression for X:
X = (1 + 25Y)/18 -> X = (1 + 25(-25)/13)/18 -> X = 26/13
Therefore, the solution to both equations (the intersection point) is (X, Y) = (26/13, -25/13).
To review the graph of a system of two linear equations, we can plot the equations on a coordinate plane and look for their intersection point.
1. Let's plot the first equation: X - 5Y = 7.
To graph this equation, we need to solve it for Y in terms of X:
-5Y = -X + 7
Y = (1/5)X - (7/5)
Now we can choose some X-values and calculate the corresponding Y-values:
X = 0, Y = -7/5
X = 5, Y = 0
X = 10, Y = 1
Plotting these points on the coordinate plane, we get a line:
|
2 |
1 | . (10, 1)
0 | . (5, 0)
_____|_____________________
-5 | .
-6 | .
-7 | . (0, -7/5)
|
|
|
0
|
2. Now, let's plot the second equation: 3X - 2Y = -4.
Solving this equation for Y:
-2Y = -3X - 4
Y = (3/2)X + 2
Choosing some X-values:
X = 0, Y = 2
X = -1, Y = -1/2
X = 4, Y = 8
Plotting these points:
|
10 |
8 | . (4, 8)
6 | .
4 | .
2 | .
0 | . (0, 2)
-2 | . (2/3, 0)
-4 | . (-4/3,-1/2)
_____|_____________________
-10 |
3. The intersection point is where the two lines intersect. From the graph, it appears to be approximately (2/3, 0).
Therefore, the solution to the system of equations is X = 2/3 and Y = 0, or (2/3, 0) as the intersection point.
To find the intersection point of the two equations, we need to solve the system of equations by using either the substitution method or the elimination method.
Let's solve the system of equations using the elimination method:
1. Multiply the first equation by 3 and the second equation by 1, such that the coefficients of the x term will be the same in both equations:
Equation 1: 3(X - 5Y) = 3(7) -> 3X - 15Y = 21
Equation 2: 1(3X - 2Y) = 1(-4) -> 3X - 2Y = -4
2. Subtract the resulting equations to eliminate the x terms:
(3X - 15Y) - (3X - 2Y) = 21 - (-4)
3X - 15Y - 3X + 2Y = 21 + 4
-13Y = 25
3. Solve for Y by dividing both sides of the equation by -13:
-13Y / -13 = 25 / -13
Y = -25/13
4. Substitute the value of Y back into one of the original equations to solve for X. Let's use the first equation:
X - 5(-25/13) = 7
X + 125/13 = 7
X = 7 - 125/13
X = 91/13 - 125/13
X = -34/13
So, the solution to the system of equations is X = -34/13 and Y = -25/13.
Therefore, the intersection point of the two equations is (-34/13, -25/13) on the graph.