Question

2) 5|x| - 3|x| = 33; 4) 42 – 63|x| = 65Jx).2) 100 - |x| = -49|x| + 124;4) 29/х - |x| - 13 = - 22|x|.​

Answer 1

Ответ:

Пошаговое объяснение:

2) 2|x|=33,  |x|=16,5,  x=16,5  или  x=-16,5

4) 42=63|x|+65|x|,  42=128|x|,  |x|=42/128,   |x|=21/64,  x=21/64  или  x= -21/64

2) 49|x|-|x|=124-100,  48|x|=24,  |x|=24/48=1/2,   x=1/2  или x=-1/2

Answer 2

Let's solve each equation step by step:

1. 5|x| - 3|x| = 33

Combining like terms on the left side, we get:

2|x| = 33

Dividing both sides by 2, we have:

|x| = 16.5

Since the absolute value of a number is always positive, we can rewrite this as:

x = 16.5 or x = -16.5

2. 100 - |x| = -49|x| + 124

First, let's simplify the right side:

-49|x| + 124 = -49|x| + 124

Next, we can separate the equation into two cases based on the sign of x.

When x is positive:

100 - x = -49x + 124

Simplifying further:

50x = 24

x = 24/50 = 12/25

When x is negative:

100 + x = 49x + 124

Simplifying further:

50x = -24

x = -24/50 = -12/25

So the solutions are x = 12/25 and x = -12/25.

3. 42 – 63|x| = 65Jx|

It seems there is a typo in the equation. The term "65Jx|" is unclear. If you provide the correct equation, I can help you solve it.

4. 29/х - |x| - 13 = - 22|x|

Let's simplify the left side first:

29/x - |x| - 13 = -22|x|

Multiplying the equation by x to eliminate the fraction:

29 - x|x| - 13x = -22x

Next, we can separate the equation into two cases based on the sign of x.

When x is positive:

29 - x^2 - 13x = -22x

Simplifying further:

x^2 - 11x + 29 = 0

This quadratic equation does not have real solutions. The discriminant (b^2 - 4ac) is negative.

When x is negative:

29 - (-x)^2 - 13(-x) = -22(-x)

Simplifying further:

x^2 - 11x + 29 = 0

This is the same quadratic equation as before and does not have real solutions.

Therefore, there are no real solutions to this equation.