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3 years ago

Which of the graphed systems share the same solution set?

Mathematics

2 answers:

aivan3 [116]3 years ago

8 0

Graph 3 ///////////////

timurjin [86]3 years ago

5 0

The third graphic system shares the same set of solutions.
Notice that the first graph is two parallel lines that never cut.
The second graph has two lines that intersect at a point. Therefore, compare a single solution.
The third graph shares the whole set of solutions since, for each x, both lines take the same Y value
answer
Third graphic.

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Describe and correct the error(s) made in each of the problems below.

ladessa [460]

Answer:

$\displaystyle \frac{1-x}{(5-x)(-x)} =-\frac{x-1 }{ x(x-5)}$

$\displaystyle \frac{5}{s}\times \frac{2}{5} =\frac{2}{s}$

Step-by-step explanation:

<u>Errors in Algebraic Operations </u>

It's usual that students make mistakes when misunderstanding the application of algebra's basic rules. Here we have two of them

When we change the signs of all the terms of a polynomial, the expression must be preceded by a negative sign
When multiplying negative and positive quantities, if the number of negatives is odd, the result is negative. If the number of negatives is even, the result is positive.
Not to confuse product of fractions with the sum of fractions. Rules are quite different

The first expression is

$1-x / (5-x)(-x)=x-1 / x(x-5)$

Let's arrange into format:

$\displaystyle \frac{1-x}{(5-x)(-x)} =\frac{x-1 }{ x(x-5)}$

We can clearly see in all of the factors in the expression the signs were changed correctly, but the result should have been preceeded with a negative sign, because it makes 3 (odd number) negatives, resulting in a negative expression. The correct form is

$\displaystyle \frac{1-x}{(5-x)(-x)} =-\frac{x-1 }{ x(x-5)}$

Now for the second expression

$5/s+2/5=2/s$

Let's arrange into format

$\displaystyle \frac{5}{s}+\frac{2}{5} =\frac{2}{s}$

It's a clear mistake because it was asssumed a product of fractions instead of a SUM of fractions. If the result was correct, then the expression should have been

$\displaystyle \frac{5}{s}\times \frac{2}{5} =\frac{2}{s}$

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