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3 years ago

-7x + y = -19 –2x+3y=-19

Mathematics

1 answer:

ZanzabumX [31]3 years ago

6 0

Answer:x=3,y=-5

Step-by-step explanation:

-7x+y=-19...(1)

-2x+3y=-19...(2)

Subject of formula from (1)

Y=-19+7x....(3)

Substitute (3) into (2)

-2x+3(-19+7x)=-19

-2x-57+21x=-19

Collect like terms

-2x+21x=-19+57

19x=38

X=2

Substitute x=2 to equation (3)

Y=-19+7(2)

Y=-19+14

Y=-5

X=2 and y=-5

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3 years ago

Which equation represents the line that passes through the points (-5,2) and (9,0)

V125BC [204]

Step-by-step explanation:

Given points 

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First calculating slope(m)

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3 years ago

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3 years ago

There are premium tickets costing $99 each and rest are regualr tickets costing $25 each. A pile of 120 tickets have a total val

polet [3.4K]

Answer:

There are 38 premium tickets and 82 regular tickets in a pile.

Step-by-step explanation:

Given,

Total number of tickets = 120

Total amount = $5812

Solution,

Let the number of premium tickets be x.

And the number of regular tickets be y.

Total number of tickets is the sum of total number of premium tickets and total number of regular tickets.

So the equation can be written as;

$x+y=120\ \ \ \ \ equation\ 1$

Again,total amount is the sum of total number of premium tickets multiplied by cost of each premium ticket and total number of regular tickets multiplied by cost of each regular ticket.

So the equation can be written as;

$99x+25y=5812\ \ \ \ equation\ 2$

Now We will multiply equation 1 by 25 we get;

$x+y=120\\25(x+y)=120\times25\\25x+25y= 3000 \ \ \ \ equation\ 3$

Now Subtracting equation 3 from equation 2 we get;

$(99x+25y)-(25x+25y)=5812-3000\\\\99x+25y-25x-25y=2812\\\\74x=2812\\\\x=\frac{2812}{74}=38$

We will now substitute the value of x in equation 1 we get;

$x+y=120\\38+y=120\\y=120-38\\y=82$

Hence There are 38 premium tickets and 82 regular tickets in a pile.

3 0

3 years ago

2. what value of b makes the following equation true

Svetradugi [14.3K]

Answer:

2. 10

3. D. 1 for all n

Step-by-step explanation:

2. The applicable rules of exponents are ...

(a^b)(a^c) = a^(b+c)

a^b = 1/a^-b

$\dfrac{4^8}{(4^2)^{-3}}\div 4^4=\dfrac{4^8}{4^{-6}\cdot4^4}=\dfrac{4^8}{4^{-2}}\\\\=4^8\cdot4^2=4^{10}$

The value of n is 10.

3. Using the above rules of exponents, the expression simplifies to ...

6^(-n+n) = 6^0 = 1

The value is 1 for any n.

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3 years ago