Answer:
II. One and only one solution
Step-by-step explanation:
Determine all possibilities for the solution set of a system of 2 equations in 2 unknowns. I. No solutions whatsoever. II. One and only one solution. III. Many solutions.
Let assume the equation is given as;
x + 3y = 11 .... 1
x - y = -1 ....2
Using elimination method
Subtract equation 1 from 2
(x-x) + 3y-y = 11-(-1)
0+2y = 11+1
2y = 12
y = 12/2
y = 6
Substitute y = 6 into equation 2:
x-y = -1
x - 6 = -1
x = -1 + 6
x = 5
Hence the solution (x, y) is (5, 6)
<em>Hence we can say the equation has One and only one solution since we have just a value for x and y</em>
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Using a calculator I got 56.36
The Solution for the given system of equations is (-2) and
.
The given are linear equations as follow:
y =
+ 3 .. ... ...(1)
and x = -2. .. .... ...(2)
We already know the first part of the solution (x) which is -2. We can find the other part (y) by putting the value of equation (2) in equation (1).
By putting the values of x in equation (1), we get
y =
+ 3
y =
+ 3
Taking the L. C. M of denominators which will be '3', we get:
y = 
y = 
So the second part (y) of the solution of the given equation is
.
Hence, the overall solution to the given system of equation is
.
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Answer:
1. Below 2. B. d = 8h
Step-by-step explanation:
The graph is a bit confusing. But basically, substitute x for the numbers they give you.
5x
5 · 1 = 5 = y
5 · 2 = 10 = y
5 · 3 = 15 = y
5 · 4 = 20 = y
7x
7 · 1 = 7 = y
7 · 2 = 14 = y
7 · 3 = 21 = y
7 · 4 = 28 = y
Answer:
14 :)
Step-by-step explanation: