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3 years ago

I'd really appreciate some help on these questions, I'm really struggling

Mathematics

1 answer:

nikklg [1K]3 years ago

6 0

We can use linear combinations of the equations to eliminate variables.

3x - 4y = 1
-2x + 3y = 1

To eliminate y we'll make the linear combination of 3 times the first equation minus four times the second.

9x - 12y = 3
-8x + 12y = 4

Adding,
x = 7

We could solve for y directly but let's use another linear combination, twice the first plus three times the second:

2(3x - 4y) + 3(-2x + 3y)= 2(1)+3(1)

y = 5

Check: 3(7)-4(5)=1 good. -2(7)+3(5)=1 good.

Q18 Answer: (7,5)

y = -3x + 5
5x - 4y = -3

4y +1(5x - 4y) = 4(-3x + 5) + 1(-3)

5x = -12x + 20 -3

17 x = 17

x = 1
y = -3(1) + 5 = 2
Check: 5(1) - 4(2) = -3 good

Q19 Answer (1,2)

6x + 5y = 25
x = 2y + 24
6x = 12y + 144
5y = 25 - 12y - 144
17y = -119
y = -119/17= -7
x = 2y+24= 10
Check: 6(10)+5(-7)=25 good 2y+24=2(-7)+24=10=x good

Q20 Answer (10,-7)

3x + y = 18
-7x + 3y = -10
9x + 3y = 54
9x - -7x = 54 - -10
16x = 64
x=4
y = 18 -3x = 18-12=6

Check: 3(4)+6=18 good, -7(4)+3(6)=-10 good

Q21 Answer: (4,6)

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Let's call the aces $h, d, c, s$ for hearts, diamonds, clubs and spades. So, $h, d$ are red and [ted] c, s[/tex] are black.

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Similarly, the probability of drawing a red ace first and then a black ace is represented by the following couples:

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