Answer:
Ok, we have a system of equations:
6*x + 3*y = 6*x*y
2*x + 4*y = 5*x*y
First, we want to isolate one of the variables,
As we have almost the same expression (x*y) in the right side of both equations, we can see the quotient between the two equations:
(6*x + 3*y)/(2*x + 4*y) = 6/5
now we isolate one off the variables:
6*x + 3*y = (6/5)*(2*x + 4*y) = (12/5)*x + (24/5)*y
x*(6 - 12/5) = y*(24/5 - 3)
x = y*(24/5 - 3)/(6 - 12/5) = 0.5*y
Now we can replace it in the first equation:
6*x + 3*y = 6*x*y
6*(0.5*y) + 3*y = 6*(0.5*y)*y
3*y + 3*y = 3*y^2
3*y^2 - 6*y = 0
Now we can find the solutions of that quadratic equation as:

So we have two solutions
y = 0
y = 2.
Suppose that we select the solution y = 0
Then, using one of the equations we can find the value of x:
2*x + 4*0 = 5*x*0
2*x = 0
x = 0
(0, 0) is a solution
if we select the other solution, y = 2.
2*x + 4*2 = 5*x*2
2*x + 8 = 10*x
8 = (10 - 2)*x = 8x
x = 1.
(1, 2) is other solution
Good morning ☕️
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Answer:
x=4
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Step-by-step explanation:
-Solving:
2x + 5 – 3 (2x - 7) = 2 (3x – 5) – 4
2x + 5 – 6x + 21 = 6x – 10 – 4
-4x + 26 = 6x - 14
10x= 26+14
10x=40
x=4
-Checking:
x=4
2(4) + 5 – 3 (2(4) - 7) = 8+5-3(8-7) = 13-3(1) =13-3=10
2 (3(4) – 5) – 4 = 2(12-5)-4 = 2(7)-4=14-4=10.
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:)
Solution:
Given, x=-3, y=6, and z=4
Putting these values in the given equation:
-15+(-x)+y
= -15+(-(-3)+6
= -15+3+6
= -15+9
= -6
Answer:
C. -3, 3, -3, 3, ...
Step-by-step explanation:
-3×-3 = 3
3×-3 = -3
etc...
Hope this helps! :)
The 2 angles at a vertex are supplementary (one interior and one exterior) the exterior angle is = the the sum of the 2 remote interior angles..