Cumulative Assessment
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Solve each inequality separately.
| At Least One Integer Solution | No Integer Solutions |
|---|---|
| 5x-6+x≥2x-8 9x-3<12 or 6x+2>-10 5(x-1)≤5x-3 |
2(3x+8)>3(2x+6) 17<4x+5<21 x-8+4x ≤ 3(x-3)+2x |
We are asked to place each inequality in the correct box. Let's begin by naming each one of the inequalities. A)& 5x-6+x≥2x-8 D) x-8 +4x ≤ 3(x-3) + 2x B)& 2(3x+8) > 3(2x+6) E) 9x-3 < 12 or 6x+2>-10 C)& 17<4x+5<21 F) 5(x-1)≤5x-3 Let's solve each of them separately!
LHS+6≥RHS+6
Add terms
LHS-2x≥RHS-2x
.LHS /4.≥.RHS /4.
a/b=.a /2./.b /2.
All numbers greater than - 12 are solutions to the inequality. Therefore, any integer greater than this number is a solution, such as x=1 or x=2.
Subtract 5 from each expression
Divide each expression by 4
orcompound inequality, we will solve each of the inequalities separately and then combine the solutions.
orinequality we can already tell that it has at least one integer solution. For example, x=1 satisfies the first inequality and this is enough.
Distribute 5
LHS-5x≤RHS-5x
Flip inequality and change signs
Finally, we can fill in the table with the information we have found.
| At Least One Integer Solution | No Integer Solutions |
|---|---|
| 5x-6+x≥2x-8 9x-3<12 or 6x+2>-10 5(x-1)≤5x-3 |
2(3x+8)>3(2x+6) 17<4x+5<21 x-8+4x ≤ 3(x-3)+2x |