Question

Mathematics MH helpo

Answer 1

The inequalities that require flipping the sign are:

-5x - 10 > 5

$\frac{x}{-7} + 3 \leq 4$

Solution:

Let us the inequalities one by one

You can perform on operations on both sides of inequality, and have its truth value unchanged

But if we multiply or divide by a negative number , we must flip the sign

option 1)

-5x - 10 > 5

Move -10 from L.H.S to R.H.S

-5x > 5 + 10

-5x > 15

Divide the above expression by 5

$-x > 3$

Divide the above inequality by -1, so we must flip the sign

x < -3

option 2)

$7x - 5 \leq 16$

Move the constant term from L.H.S to R.H.S

$7x \leq 16 + 5\\\\7x \leq 21$

Divide the above inequality by 7

$x \leq 3$

This does not required flipping the symbol

option 3

$\frac{x}{5} - 6 > -11$

Move the constant term from L.H.S to R.H.S

$\frac{x}{5} > -11 + 6\\\\\frac{x}{5} > -5$

Multiply both the sides by 5

$x > -25$

This does not required flipping the symbol

option 4

$x + 12 \leq 29$

Move the constant term from L.H.S to R.H.S

$x \leq 29 - 12\\\\x \leq 17$

This does not required flipping the symbol

option 5

$\frac{x}{-7} + 3 \leq 4$

Move the constant term from L.H.S to R.H.S

$\frac{x}{-7} \leq 4-3\\\\\frac{x}{-7} \leq 1$

Multiply both the sides by -7, so we must flip the sign

$x \geq -7$

Thus this requires flipping the sign

Answer 2

Answer: -5x - 10 > 5

Step-by-step explanation: