How do you solve the inequality #d + 6 ≤ 4d - 9# or #3d -1 <2d + 4#?

add 9 to both sides

Divide by 3

Add 1 to both sides

To solve the inequality ( d + 6 \leq 4d - 9 ) or ( 3d - 1 < 2d + 4 ), we solve each inequality separately and then combine the solutions.

For the first inequality: [ d + 6 \leq 4d - 9 ]

Subtract ( d ) from both sides: [ 6 \leq 3d - 9 ]

Add 9 to both sides: [ 15 \leq 3d ]

Divide both sides by 3: [ 5 \leq d ]

For the second inequality: [ 3d - 1 < 2d + 4 ]

Subtract ( 2d ) from both sides: [ d - 1 < 4 ]

Add 1 to both sides: [ d < 5 ]

Now, to combine the solutions: [ d \geq 5 ] from the first inequality, [ d < 5 ] from the second inequality.

So, the solution to the given compound inequality is ( d \in (-\infty, 5) ).

To solve the inequality (d + 6 \leq 4d - 9) or (3d - 1 < 2d + 4), we need to solve each inequality separately and then consider the solutions that satisfy either one of them.

1. Solve (d + 6 \leq 4d - 9): [d + 6 \leq 4d - 9] Subtract (d) from both sides: [6 \leq 3d - 9] Add 9 to both sides: [15 \leq 3d] Divide both sides by 3 (since 3 is positive, the direction of the inequality remains the same): [5 \leq d]

2. Solve (3d - 1 < 2d + 4): [3d - 1 < 2d + 4] Subtract (2d) from both sides: [d - 1 < 4] Add 1 to both sides: [d < 5]

Now, we need to consider the solutions to either inequality:

For the first inequality, (5 \leq d), the solutions are all real numbers greater than or equal to 5.

For the second inequality, (d < 5), the solutions are all real numbers less than 5.

Therefore, the combined solution to the given compound inequality is (d < 5) or (d \geq 5), which means all real numbers except 5.