From the first equation,
x+5 = 3(y+5)
x = 3y + 15 - 5
Now substitue x in the second equation with (3y +15 - 5).
x-5 = 7(y-5)
(3y+15-5) - 5 = 7(y-5)
3y +5 = 7y - 35
-4y = - 40
y = 10
Since y is 10, and x is (3y +15 - 5),
x = 30 + 15 - 5 = 40
Can you please explain? Or maybe continue your question please?
Answer:
Step-by-step explanation:
Isolate d
.
Divide both sides of the equation by b
Simplify
For cos(2x) * (2cos(x) + 1) = 0, use the double angle identity for cos(2x), which is cos^2 x - sin^2 x = cos^2 x - (1-cos^2) = 2cos^2 x - 1.
So we have (2cos^2 x - 1)(2cos x + 1) = 0. So 2cos^2 x -1 = 0 or x = 0 and 2pi.
For 2sec^2 x + tan^2 x - 3 = 0, use the identity sec^2 x = tan^2 x + 1, so we have
2(tan^2 x + 1) + tan^2 x - 3 = 0 or
<span>2tan^2 x + tan^2 x - 1 = 0 or
</span>3 tan^2 x = 1.
So x = pi/2, pi/2 + pi = 3pi/2.
Answer:
34
Step-by-step explanation:
