Anonymous
Anonymous asked in Science & MathematicsMathematics · 3 weeks ago

Systems of equation word problem? ?

Jen butler has been pricing Speed-Pass train fares for a group trip to New York. Three adults and four children must pay \$98. Two adults and three children must pay \$69. Find the price of the adult’s ticket and the price of a child’s ticket.

Relevance
• John
Lv 4
3 weeks ago

if jen sits on my lap the whole way she rides for free. how does thst calculate in?

• Ian H
Lv 7
3 weeks ago

3a + 4c = 98 ..... r

2a + 3c = 69 ..... s

c = 3s – 2r = 11

a = (69 – 33)/2 = 18

• 3 weeks ago

y - children

3[3x + 4y = 98] eq1

4[2x + 3y = 69]  eq2

--------------------

9x + 12y = 294

-[8x + 12y = 276]

----------------------

x = 18

solving for y from eq1

3x + 4y = 98

3(18) + 4y = 98

54 + 4y = 98

4y = 98 - 54

4y = 44

y = 11

The price of the adults tickets is \$18 and price of children ticket is \$11...

• RR
Lv 7
3 weeks ago

Three adults and four children must pay \$98.

(i) 3a + 4c = 98

Two adults and three children must pay \$69.

(ii) 2a + 3c = 69

You need to eliminate one unknown by making them equal in both equations and subtracting one from the other:

(i) x 2

= (iii) 6a + 8c = 196

and (ii) x 3

= (iv) 6a + 9c = 207

subtract (iii) from (iv)

c = 11

substitute into (i)

3a + (4 x 11) = 98

3a + 44 = 98

3a = 54

a = 18

, Child = \$11

• 3 weeks ago

adults pay \$32.66, kids pay \$10.89.

basic maths 98 dived by 3 , gives you the price per adult , multiply by 2 aduts , left over divided by 3 kids gives you the price per kid.

• 3 weeks ago

3a + 4c = 98

2a + 3c = 69

6a + 8c = 196

6a + 9c = 207

c = 11, a = 18

The price of an adult’s ticket is \$18

and the price of a child’s ticket is \$11.

• sepia
Lv 7
3 weeks ago

Jen butler has been pricing Speed-Pass train fares

for a group trip to New York.

Three adults and four children must pay \$98.

Two adults and three children must pay \$69.

3a + 4c = 98

2a + 3c = 69

6a + 8c = 196

6a + 9c = 207

c = 11, a = 18

The price of an adult’s ticket is \$18

and the price of a child’s ticket is \$11.

• 3 weeks ago

3A + 4C = 98

2A +3C = 69

SUBTRACT

A + C = 29

MULTIPLY BY 3

3A + 3C = 87

SUBTRACT FROM EQUATION 1

C = 11

SUBSTITUTE VALUE OF C IN ANY EQUATION

A = 18

• 3 weeks ago

let price of adult ticket = \$ x

price of child ticket = \$ y

3 adults and 4 children must pay \$ 98

so equation becomes

3x + 4y = 98

2 adults and 3 children must pay \$ 69

equation becomes

2x + 3y = 69

solving the two equations for x and y

In this case multiply 1st equation by 2 and 2nd equation by 3

6x + 8y =196

6x + 9y =207

Subtract the 1st from the second .

y= 11

x= 18

\$ 11.00 for child

• 3 weeks ago

3a + 4c = 98

2a + 3c = 69

multiply first by 2, second by –3 and add

6a + 8c = 196

–6a – 9c = 207

–c = –11

c = 11

2a + 3c = 69

2a + 3•11 = 69

2a + 33 = 69

2a  = 36

a = 18

check

3a + 4c = 98

3•18 + 4•11 = 98

54 + 44 = 98 ok

2a + 3c = 69

2•18 + 3•11 = 69

36 + 33 = 69 ok