Solve.
y=2x - 7
3x - 4y=8
Eq1: Y = 2X - 7.
Eq2: 3X - 4Y = 8.
In Eq2, substitute 2X - 7 for Y:
3X - 4(2X - 7) = 8,
3X - 8X +28 = 8,
-5X = 8 - 28 = -20,
X = 4.
In Eq1, substitute 4 for X:
Y = 2*4 - 7 = 1.
Solution Set = (X,Y) = (4,1).
To solve the system of equations, we will use either the substitution method or the elimination method. Let's use the elimination method in this case.
Given:
1) y = 2x - 7
2) 3x - 4y = 8
We want to eliminate one variable so that we can solve for the other variable. To do this, we can multiply one or both of the equations by appropriate constants so that coefficients of one of the variables in both equations will cancel out when added or subtracted.
Let's start by multiplying equation (1) by 4, which will give us:
4y = 8x - 28 (equation 3)
Now, we can subtract equation (3) from equation (2) to eliminate the 'y' variable:
(3x - 4y) - (8x - 28) = 8
3x - 4y - 8x + 28 = 8
-5x - 4y + 28 = 8
Next, let's rearrange the equation to isolate the 'y' variable:
-5x - 4y = 8 - 28
-5x - 4y = -20 (equation 4)
Now, we have two equations:
1) y = 2x - 7
4) -5x - 4y = -20
We can eliminate the 'y' variable by multiplying equation (1) by 4 and adding it to equation (4):
4y = 8x - 28 (equation 5)
-5x - 4y = -20 (equation 4)
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-5x + 8x - 28 = -20
Simplifying the equation:
3x - 28 = -20
Next, add 28 to both sides of the equation:
3x - 28 + 28 = -20 + 28
3x = 8
Finally, divide both sides of the equation by 3 to solve for 'x':
3x/3 = 8/3
x = 8/3
Now that we have found the value of 'x', we can substitute it back into equation (1) to find the value of 'y':
y = 2 * (8/3) - 7
y = 16/3 - 7
y = 16/3 - 21/3
y = -5/3
Therefore, the solution to the system of equations is x = 8/3 and y = -5/3.