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notsponge [240]
3 years ago
9

Find all solutions of the given system of equations and check your answer graphically. HINT [See Examples 1-4.] (If there is no

solution, enter NO SOLUTION. If the system is dependent, express your answer in terms of x, where y = y(x).)
3x − 4y= 4
9x − 12y= 12
Mathematics
1 answer:
Likurg_2 [28]3 years ago
3 0

Answer:

Infinitely\ many\ solutions\ exist.\\\\Solutions\ are\ (x,\frac{3}{4}x-1)

Step-by-step explanation:

Given\ equations\ are\\\\3x-4y=4.................eq(1)\\\\9x-12y=12..............eq(2)\\\\divide\ eq(2)\ by\ 3\\\\\frac{1}{3}(9x-12y=12)\\\\\Rightarrow 3x-4y=4\\\\Hence\ equations\ represent\ the\ same\ line.\\Hence\ Infinitely\ many\ solutions\ exist.\\\\3x-4y=4\\\\4y=3x-4\\\\y=\frac{3}{4}x-1\\\\Solutions\ are\ (x,\frac{3}{4}x-1)

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Can you help me with some steps to approach this proof .
Elis [28]

To prove two sets are equal, you have to show they are both subsets of one another.

• <em>X</em> ∩ (⋃ ) = ⋃ {<em>X</em> ∩ <em>S</em> | <em>S</em> ∈ }

Let <em>x</em> ∈ <em>X</em> ∩ (⋃ ). Then <em>x</em> ∈ <em>X</em> and <em>x</em> ∈ ⋃ . The latter means that <em>x</em> ∈ <em>S</em> for an arbitrary set <em>S</em> ∈ . So <em>x</em> ∈ <em>X</em> and <em>x</em> ∈ <em>S</em>, meaning <em>x</em> ∈ <em>X</em> ∩ <em>S</em>. That is enough to say that <em>x</em> ∈ ⋃ {<em>X</em> ∩ <em>S</em> | <em>S</em> ∈ }. So <em>X</em> ∩ (⋃ ) ⊆ ⋃ {<em>X</em> ∩ <em>S</em> | <em>S</em> ∈ }.

For the other direction, the proof is essentially the reverse. Let <em>x</em> ∈ ⋃ {<em>X</em> ∩ <em>S</em> | <em>S</em> ∈ }. Then <em>x</em> ∈ <em>X</em> ∩ <em>S</em> for some <em>S</em> ∈ , so that <em>x</em> ∈ <em>X</em> and <em>x</em> ∈ <em>S</em>. Because <em>x</em> ∈ <em>S</em> and <em>S</em> ∈ , we have that <em>x</em> ∈ ⋃ , and so <em>x</em> ∈ <em>X</em> ∩ (⋃ ). So ⋃ {<em>X</em> ∩ <em>S</em> | <em>S</em> ∈ } ⊆ <em>X</em> ∩ (⋃ ).

QED

• <em>X</em> ∪ (⋂ ) = ⋂ {<em>X</em> ∪ <em>S</em> | <em>S</em> ∈ }

Let <em>x</em> ∈ <em>X</em> ∪ (⋂ ). Then <em>x</em> ∈ <em>X</em> or <em>x</em> ∈ ⋂ . If <em>x</em> ∈ <em>X</em>, we're done because that would guarantee <em>x</em> ∈ <em>X</em> ∪ <em>S</em> for any set <em>S</em>, and hence <em>x</em> would belong to the intersection. If <em>x</em> ∈ ⋂ , then <em>x</em> ∈ <em>S</em> for all <em>S</em> ∈ , so that <em>x</em> ∈ <em>X</em> ∪ <em>S</em> for all <em>S</em>, and hence <em>x</em> is in the intersection. Therefore <em>X</em> ∪ (⋂ ) ⊆ ⋂ {<em>X</em> ∪ <em>S</em> | <em>S</em> ∈ }.

For the opposite direction, let <em>x</em> ∈ ⋂ {<em>X</em> ∪ <em>S</em> | <em>S</em> ∈ }. Then <em>x </em>∈ <em>X</em> ∪ <em>S</em> for all <em>S</em> ∈ . So <em>x</em> ∈ <em>X</em> or <em>x</em> ∈ <em>S</em> for all <em>S</em>. If <em>x</em> ∈ <em>X</em>, we're done. If <em>x</em> ∈ <em>S</em> for all <em>S</em> ∈ , then <em>x</em> ∈ ⋂ , and we're done. So ⋂ {<em>X</em> ∪ <em>S</em> | <em>S</em> ∈ } ⊆ <em>X</em> ∪ (⋂ ).

QED

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Answer:

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Step-by-step explanation:

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<em><u>4 pints of strawberries and 2 pints of blueberries are bought</u></em>

<em><u>Solution:</u></em>

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Cost per pint of strawberry = $ 1.60

Cost per pint of blueberry = $ 2.30

<em><u>A shopper bought twice as many pints of strawberries as pints of blueberries</u></em>

Therefore,

a = 2b --------- eqn 1

<em><u>They spent a total of $11.00. Therefore we frame a equation as:</u></em>

pints of strawberries bought x Cost per pint of strawberry + pints of blueberries cost x Cost per pint of blueberry = 11

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<em><u>Substitute eqn 1 in eqn 2</u></em>

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5.5b = 11

Divide both sides by 11

b = 2

<em><u>Substitute b = 2 in eqn 1</u></em>

a = 2(2)

a = 4

Thus 4 pints of strawberries and 2 pints of blueberries are bought

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