Question

Harry had $32. He spent all the money buying three notebooks for x dollars each and four packs of index cards for y dollars each. If Harry had bought five notebooks and five packs of index cards, he would have run short by $18. The following system of equations models this scenario: 3x 4y = 32 5x 5y = 50 Use the system of equations to solve for x and y. (8, 2) (1, 5) (2, 8) (5, 1).

Answer 1

x = notebooks

y = indexboks

3x+4y=32 ( here is spent all money so I add 32 here )

5x+5y=50

by elemination method

3x+4y=32×5

5x+5y=50×4 ( here I'm going to eliminate y )

15 x+20y= 160

20x+20y= 200

- - = -

____________

-5x = -40

x = - 40 / -5

x = 8

here - × - = + so iam taking positive 8

now substitute the value of x in any equation

3×8+4y = 32

24 + 4y = 32

4y = 32-24

y = 8 / 4

y = 2

CHECKING

SUBSTITUTE BOTH VALUES IN ANY EQUATION

5×8+5×2 = 50

40 + 10 = 50

50 = 50

I SUBSTITUTE IN 2ND EQUATION

Answer 2

(8,2)

Explanation:

Given that:

3x + 4y = 32 --------- eq1

5x + 5y = 50 --------- eq2

From taking 5 common from eq2:

x + y = 10

Or it can also be written as:

x = 10 - y ----------eq3

Now put this value of x in eq1

3(10 - y) + 4y = 32

By simplifying:

30 - 3y +4y= 32

30 +y = 32

Subtracting 30 from both sides:

y = 32 - 30

y = 2

Putting value of y in eq 3

x = 10 - 2

x = 8