Question 148164

{{{(2x+7)(3x+2)<=0}}}  Factor the left side

{{{(2x+7)(3x+2)=0}}} Set the left side equal to zero

Set each individual factor equal to zero:

{{{2x+7=0}}} or  {{{3x+2=0}}}

Solve for x in each case:

{{{x=-7/2}}} or  {{{x=-2/3}}}

So our critical values are {{{x=-7/2}}} and  {{{x=-2/3}}}

Now set up a number line and plot the critical values on the number line

{{{number_line( 600, -10, 10,-7/2,-2/3)}}}

So let's pick some test points that are near the critical values and evaluate them.

Let's pick a test value that is less than {{{-7/2}}} (notice how it's to the left of the leftmost endpoint):

So let's pick {{{x=-4}}}

{{{(2(-4)+7)(3(-4)+2)<= 0}}} Plug in {{{x=-4}}}

{{{10<= 0}}} Evaluate and simplify the left side

Since the inequality is false, this means that the interval does <b>not</b> work. So this interval is <b>not</b> in our solution set and we can ignore it.

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Let's pick a test value that is in between {{{-7/2}}} and {{{-2/3}}}:

So let's pick {{{x=-2}}}

{{{(2(-2)+7)(3(-2)+2)<= 0}}} Plug in {{{x=-2}}}

{{{-12<= 0}}} Evaluate and simplify the left side

Since the inequality is true, this means that the interval works. So this tells us that this interval is in our solution set.

So part our solution in interval notation is <font size="8">[</font>*[Tex \LARGE -\frac{7}{2},-\frac{2}{3}]<font size="8">]</font>

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Let's pick a test value that is greater than {{{-2/3}}} (notice how it's to the right of the rightmost endpoint):

So let's pick {{{x=1}}}

{{{(2(1)+7)(3(1)+2)<= 0}}} Plug in {{{x=1}}}

{{{45<= 0}}} Evaluate and simplify the left side

Since the inequality is false, this means that the interval does <b>not</b> work. So this interval is <b>not</b> in our solution set and we can ignore it.

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Summary:

So the solution in interval notation is:

<font size="8">[</font>*[Tex \LARGE -\frac{7}{2},-\frac{2}{3}]<font size="8">]</font>

Here's a graph to prove it

{{{ graph( 500, 500, -10, 10, -10, 10, (2x+7)(3x+2)) }}}