−4x + y = 6 −5x − y = 21
1 answer:
You have in the first equation "+y" and in the second "-y"
So, you add the 2 equations, so we can remove this 2 letters since they're opposite numbers.
So, we get:
-4x + y -5x - y = 21 + 6
-4x -5x = 21 + 6
Then, you add
-4x -5x = 21 + 6
-9x = 27
Then, you move the -9 to the other part, next to the 27. And since it was a multiplication, it becomes a division.
x =
Then, you choose an equation of the system, and you replace x by it's value so you can find y. I'll take the first equation.
-4x + y = 6
-4*-3 + y = 6
12 + y = 6
Then, you move 12 to the other part next to the 6. And since it was a positive number, it becomes a negative number.
So
y = 6 - 12
y = -6
So, the solution of the proposed system <span>is (x;y) = (6;-6)
x = 6
y = -6
Hope this Helps :)
</span>
So, you add the 2 equations, so we can remove this 2 letters since they're opposite numbers.
So, we get:
-4x + y -5x - y = 21 + 6
-4x -5x = 21 + 6
Then, you add
-4x -5x = 21 + 6
-9x = 27
Then, you move the -9 to the other part, next to the 27. And since it was a multiplication, it becomes a division.
x =
Then, you choose an equation of the system, and you replace x by it's value so you can find y. I'll take the first equation.
-4x + y = 6
-4*-3 + y = 6
12 + y = 6
Then, you move 12 to the other part next to the 6. And since it was a positive number, it becomes a negative number.
So
y = 6 - 12
y = -6
So, the solution of the proposed system <span>is (x;y) = (6;-6)
x = 6
y = -6
Hope this Helps :)
</span>
You might be interested in
This problem is a problem using
d = r * t
First part of the trip
d = 96 miles
r = ??
t = ??
Second part of the trip
d = 15 miles
r = r - 5
t = 4 - t
Solve for r first.
96/r + 15/(r - 5) = 4 hours multiply through by r and r - 5
96(r - 5) + 15r = 4*(r - 5)(r) Remove the brackets on the left
96r - 480 + 15r = 4*r*(r - 5) Collect like terms on the left.
111r - 480 = 4r(r - 5) Remove the brackets on the right.
111r - 480 = 4r^2 - 20r Bring the left side over to the right side.
0 = 4r^2 - 20r - 111r + 480 Collect like terms.
0 = 4r^2 - 131r + 480
The only way I could do this was to use the quadratic equation.
x = (131 + 97.38)/8
x = 28.55
x = (131 - 97.37)/8
x = 4.20
4.2 isn't going to work because the slower speed has to be larger than 5, otherwise you will get a negative speed (as this solution will give), so 4.20 is an extraneous solution.
r = 28.55 for the first part of the trip and
r = 23.55 for the second part of the trip.
Check.
Let's see if the times add up to 4
d/r = t
96/28.55 = 3.36
15/22.75 = 0.64
The times do = 4, so the rates are correct.
Answers
r1 = 28.55 for the 96 mile trip.
r2 = 23.55 for the 15 mile trip.
d = r * t
First part of the trip
d = 96 miles
r = ??
t = ??
Second part of the trip
d = 15 miles
r = r - 5
t = 4 - t
Solve for r first.
96/r + 15/(r - 5) = 4 hours multiply through by r and r - 5
96(r - 5) + 15r = 4*(r - 5)(r) Remove the brackets on the left
96r - 480 + 15r = 4*r*(r - 5) Collect like terms on the left.
111r - 480 = 4r(r - 5) Remove the brackets on the right.
111r - 480 = 4r^2 - 20r Bring the left side over to the right side.
0 = 4r^2 - 20r - 111r + 480 Collect like terms.
0 = 4r^2 - 131r + 480
The only way I could do this was to use the quadratic equation.
a = 4
b = - 131
c = 480
x = (131 + 97.38)/8
x = 28.55
x = (131 - 97.37)/8
x = 4.20
4.2 isn't going to work because the slower speed has to be larger than 5, otherwise you will get a negative speed (as this solution will give), so 4.20 is an extraneous solution.
r = 28.55 for the first part of the trip and
r = 23.55 for the second part of the trip.
Check.
Let's see if the times add up to 4
d/r = t
96/28.55 = 3.36
15/22.75 = 0.64
The times do = 4, so the rates are correct.
Answers
r1 = 28.55 for the 96 mile trip.
r2 = 23.55 for the 15 mile trip.