write a system of linear equations that has only one solution. solve the system of equations in two ways.

1 answer:

7 0

X + 5y = 1 . . . . (1)
8x - 2y = 3 . . . .(2)
Solution 1.
From (1), x = 1 - 5y
substituting for x in (2), we have:
8(1 - 5y) - 2y = 3
8 - 40y - 2y = 3
8 - 42y = 3
42y = 5
y = 5/42
x = 1 - 5(5/42) = 1 - 25/42 = 17/42.

Solution 2.
Multiply (1) by 8, to get:
8x + 40y = 8 . . . . . (3)
(3) - (2) = 40y - (-2y) = 8 - 3
42y = 5
y = 5/42
substitute for y into (3), to get:
8x + 40(5/42) = 8
8x + 100/21 = 8
8x = 8 - 100/21 = 68/21
x = (68/21)/8 = 17/42

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Inverse of f(x)= 2^3x + 1

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To find the inverse of a function:
1. Replace f(x) with y.
2. Switch x and y.
3. Solve for y.
4. Replace y with f-1(x) (notation for inverse fucntions)

$f(x)=2^{3x}+1$
Let's use y instead of f(x).
$y=2^{3x}+1$
Switch the variables...
$x=2^{3y}+1$
Now, we solve for y.
$x-1=2^{3y}$
$3y =log_2(x-1)$
$y=\frac{log_2(x-1)}{3}$
Now, just put that inverse function notation on there.

$\boxed{f^{-1}(x)=\frac{log_2(x-1)}{3}}$