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igor_vitrenko [27]
2 years ago
6

PLEASE I NEED THE ANSWER QUICKLY

Mathematics
2 answers:
Pavlova-9 [17]2 years ago
7 0

Answer:

x = 12 half page advertisements

y = 8 full page advertisements

Step-by-step explanation:

42x + 86y = 1192

x + y = 20

y = 20 - x

Input (20-x) as y into the first equation

42x + 86(20-x)=1192

42x + 1720 - 86x = 1192

(42x - 86x) 1720 -1720= 1192-1720

-44x = -528

x = 12

12 + y = 20

y=8

enot [183]2 years ago
5 0

a. 'x' represents half page advertisement and 'y' represents full page advertisement.

System of equation is given by 42x + 86y = 1192 , x + y =20.

b. Number of full page advertisement  Kathryn purchase is equals to 8.

c.  Number of half page advertisement  Kathryn purchase is equals to 12.

<h3>What is system of equation?</h3>

" System of equation is defined as the finite set of equations for which we find a common solution."

According to the question,

Cost of half page advertisement = $42

Cost of full page advertisement = $86

Number of advertisement Kathryn could purchase = 20

Total budget of Kathryn to purchase advertisement = $1192

'x' represents half page advertisement

'y' represents full page advertisement

As per the situation given system of equation we have,

42x + 86y = 1192                                 ______(1)

x + y = 20                        

x = 20 - y                                             _____(2)

Substitute the value of 'x' from  system of equation (2) in (1) we get,

42( 20 -y) +86y =1192\\\\\implies 840 - 42y +86y =1192\\\\\implies 44y = 1192-840\\\\\implies y = \frac{352}{44}\\ \\\implies y =8

Substitute the value of y in (2) to get 'x' ,

x=20-8

  = 12

Hence,

a. 'x' represents half page advertisement and 'y' represents full page advertisement.

System of equation is given by 42x + 86y = 1192 , x + y =20.

b. Number of full page advertisement  Kathryn purchase is equals to 8.

c.  Number of half page advertisement  Kathryn purchase is equals to 12.

Learn more about system of equation here

brainly.com/question/13997560

#SPJ3

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