Solve this inequality: 3q + 11 + 8q > 99. A. q > 8 B. q > 1/8 C. q > 88 D. q > 11

1 answer:

3 0

Hey!

To solve this problem we must first group like terms.

Original Equation :
3q + 11 + 8q > 99

New Equation {Grouped Like Terms} :
3q + 8q + 11 > 99

Now, we'll add our similar elements.

Old Equation :
3q + 8q + 11 > 99

New Equation {Added Similar Elements} :
11q + 11 > 99

Next, we'll have to change the equation. To do this we'll add minus eleven to both sides of the equation. This will give us eleven q on its own.

Old Equation :
11q + 11 > 99

New Equation {Added Subtract 11 to Both Sides} :
11q + 11 - 11 > 99 - 11

Now we simplify.

Old Equation :
11q + 11 - 11 > 99 - 11

New Equation {Old Equation Simplified} :
11q > 88

Now, we'll have to divide both sides by eleven. We do this so as to get the variable ( q ) on its own.

Old Equation :
11q > 88

New Equation {Added Divide Both Sides by 11} :
 $\frac{11q}{11}$ > $\frac{88}{11}$

Once again, we now have to simplify.

Old Equation :
$\frac{11q}{11}$ > $\frac{88}{11}$

New Equation {Old Equation Simplified} :
q > 8

So, this means that the solution to the inequality given is q > 8.

Hope this helps!

- Lindsey Frazier ♥

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Comment on the graph

I like to find the roots of higher-degree polynomials using a graphing calculator. The red curve is the cubic. Its only rational root is r=1. By dividing the function by the known factor, we have a quadratic. The graphing calculator shows its vertex, so we know immediately what the vertex form of the quadratic factor is. The linear factors are easily found from that, as we show above. (This is the "other means" we used to find the quadratic roots.)