The system of equations shown is graphed in the
x
y
–plane.
{
x
+
y
=
1
2
x
−
3
y
=
17
Determine the coordinates of the solutions of the system algebraically. Then show that your solution is valid. Show your work or explain how you found your answer.
Enter your answer and your work or explanation in the space provided. You may also use the drawing tool to help explain or support your answer.
To solve the system algebraically, we can use either substitution or elimination method.
For this problem, let's use the elimination method. We can multiply the first equation by 2 to eliminate the x term:
2(x + y) = 2(12)
2x + 2y = 24
Now we can subtract this new equation from the second equation to eliminate the x term:
12x - 3y - (2x + 2y) = 17 - 24
10x - 5y = -7
Next, divide this equation by 5 to simplify it further:
2x - y = -7/5
Now we have a simplified system of equations:
2x - y = -7/5
x + y = 12
We can now add the two equations together to eliminate the y term:
(2x - y) + (x + y) = -7/5 + 12
3x = 53/5
Dividing both sides by 3, we get:
x = 53/15
Substituting this value of x into either equation, let's use the second equation:
53/15 + y = 12
y = 12 - 53/15
y = 120/15 - 53/15
y = 67/15
So the solution to the system of equations is the point (53/15, 67/15).
To show that this solution is valid, we can substitute the values of x and y back into both original equations and check if they hold true.
For the first equation:
x + y = 1/2
53/15 + 67/15 = 1/2
120/15 = 1/2 (which simplifies to 8/10 = 1/2)
Both sides are equal, so the first equation holds true.
For the second equation:
12x - 3y = 17
12(53/15) - 3(67/15) = 17
636/15 - 201/15 = 17
435/15 = 17
29/15 = 17
The right side and left side are not equal, so the second equation does not hold true.
Since the solution (53/15, 67/15) does not satisfy the second equation, it is not a valid solution to the system of equations.