Graph the system of linear inequalities, please explain the process in solving this problem so I know how to do ones similar to
this!
1 answer:
Answer:
See attached. (Both inequalities graph to the same region.)
Step-by-step explanation:
To graph an inequality, you first replace the inequality symbol by an equal sign. Then you graph the line (or curve) described by the equation. If the inequality symbol includes the "or equal to" case, then the line (or curve) is solid and is part of the solution set. If not, then the line (or curve) is dashed and is <em>not part of the solution set</em>.
Here, the line x+y=5, will bound the solution set, but is not part of it.
Note that the line -x -y = -5 (from the second inequality) is the same ilne.
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After you have graphed the boundary of the solution region, you need to determine which side of the boundary has the solutions. I find it easy to do this by locating an x- or y-term with a positive coefficient and seeing which side of the inequality symbol that lies on. Here, both the x-term and the y-term lie on the "is greater than" side of the symbol, so the half-plane is shaded above and to the right of the line, where x- and y-values are greater than they are on the line.
In the second inequality, you can multiply the whole thing by -1 to get
... x + y > 5 . . . . same as the first inequality
or you can add (x+y) to both sides to get
... 0 < -5 +x +y
Once again, the x- and y-values that satisfy this are greater than those that are on the boundary line.
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The graph seems to show only one equation. Both are there, but the dashed blue line sits on top of the dashed red line, so you can't see it. The purple shading indicates both red and blue shading in the same area.
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Of course you graph the boundary line the way you would any linear equation. In this form, it is not too difficult to see that both the x- and y-intercepts are 5. (When x=0, y=5; and when y=0, x=5 for the boundary line's equation x+y=5.) Draw the dotted line through these intercept points, add shading above, and you have your graph.
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. Multiplying the denominator by
gives
Subtracting this from the numerator gives a remainder of
. Multiplying the denominator by
gives
and subtracting this from the previous remainder gives a new remainder of
This last remainder is exactly the same as the denominator, so divides through it exactly and leaves us with 1.
What we showed here is that
and this last expression is the quotient.
To verify this solution, we can simply multiply this by the original denominator:
which matches the original numerator.