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Svet_ta [14]
3 years ago
5

IMAGE BELOW The equations x minus 2 y = 4, 4 x + 5 y = 8, 6 x minus 5 y = 15, and x + 2 y = 0 are shown on the graph below.

Mathematics
1 answer:
alina1380 [7]3 years ago
3 0

Answer:

6x - 5y = 15 and x + 2y = 0

Step-by-step explanation:

Here we are given the following equations:

i) x-2y= 4

ii) 4x+5y= 8

iii) 6x-5y=15

iv) x+2y=0

Required:

Which system of equations has a solution of approximately (1.8, –0.9).

To find the approximate equations, substitute x and y for  1.8 and  -0.9 into all the equations respectively and check the resulting values

i) Substitute (1.8, -0.9) in x-2y= 4:

1.8 - 2(-0.9) = 4.

1.8 + 1.8 = 4.

3.6 ≠ 4.

ii) Substitute (1.8, -0.9) in 4x+5y= 8

4(1.8) + 5(-0.9) = 8

7.2 - 4.5 = 8.

2.7 ≠ 8.

iii) Substitute (1.8, -0.9) in 6x-5y=15

6(1.8) - 5(-0.9) = 15.

10.8 + 4.5 = 15

15.3 ≠ 15.

This equation has a solution that is close, therefore it is correct.

iv) Substitute (1.8, -0.9) in x+2y=0

1.8 + 2(-0.9) = 0.

1.8 - 1.8 = 0.

0 = 0.

x + 2 y = 0 has the exact value, therefore it is also correct.

The system of equations that has a solution of approximately (1.8, –0.9) are:

x+2y=0 and 6x-5y=15

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Li wants to add these items to her suitcase: a hairdryer (1.25 lb), a hand-held video game (0.6 lb), an extra video game (0.25 l
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Answer:

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Step-by-step explanation:

Li wants to add these items to her suitcase: a hairdryer (1.25 lb), a hand-held video game (0.6 lb), an extra video game (0.25 lb), and a music player (0.4 lb). Using the inequality x ≤ 2.25, can she add all these items to the suitcase? If not, what could be left out to be closest to the maximum weight allowed?

Solution:

Weight of the items

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Item equivalent to 0.25lb is the extra video game

If Li leaves out the extra game, which weighs 0.25 lb, she’d have the maximum weight allowed.

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