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olga2289 [7]
3 years ago
5

Solve algebraically the simultaneous equations

Mathematics
1 answer:
Scilla [17]3 years ago
5 0

Answer:

(5, 1) and (-5, -7)

Step-by-step explanation:

Given the simultaneous equations

x^2– y^2 = 24 .... 1

x= 3 + 2y .... 2

Substitute 2 into 1

(3-2y)^2-y^2 = 24

Expand

9 - 12y + 4y^2 - y^2 = 24

9 - 12y + 3y^2 = 24

3y^2 - 12y + 9-24 = 0

3y^2 -12y -15 = 0

y^2-4y-5 = 0

y^2+5y - y - 5 = 0

y(y+5) - 1 (y+5) = 0

y-1 = 0 and y+5 = 0

y = 1 and -5

Substitute the values of y into 2

Recall that x = 3+3y

when y = 1

x = 3 + 2(1)

x = 5

when x = -5

x = 3 + 2(-5)

x = 3 - 10

x = -7

Hence the solutions are (5, 1) and (-5, -7)

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Bottle of juice: 2.08 including tax

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What would a box and whisker plot look like for this?
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58, 62, 71, 73, 84, 89, 91, 91, 93, 97, 98, 101,104

Five number summary:
1) minimum = 58
2) 1st quartile = 72
3) median = 91
4) 3rd quartile = 98
5) maximum = 104
              __________
    --------<u>|        |         </u>|------------                     
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