Two equations:

1) 8 cards + 3 boxes = $23.54

2) 9 cards + 4 boxes = $28.67

Using variables, the two equations above becomes:

1) 8c + 3b = 23.54

2) 9c + 4b = 28.67

Solve for “b” by eliminating the “c” variable.

Multiply equation (1) by 9, multiply equation (2) by 8.

1) 72c + 27b = 211.86

2) 72c + 32 b = 229.36

Subtract equation (2) from equation (1), thus eliminating “c”.

5b = 17.50

Divide both sides by 5.

b = 3.50

Thus, ONE box of candy = $3.50.

However, we’re solving for TWO boxes of candy, so multiply $3.50 x 2.

Therefore, two boxes of candy costs $7.00!

y = cards cost

8y + 3x = 23.54

9y + 4x = 28.67

Pick one and solve for a variable. We’ll do the first eqn for X.

x = (23.54 – 8y)/3

Plug that into the 2nd eqn

9y + 4(23.54 – 8y)/3 = 28.67

9y + 31.4 – 32y/3 = 28.67

-1.6y = -2.73

y = 1.70/card

plug that in to either equation above to solve & find x to be 3.31

Then 8x+3y=23.54, and 9x+4y=28.67. Solve the first equation for y:

3y=23.54-8x

y=7.86-2.66x

Plug the value of y you just solved for back into the second equation:

9x+4(7.86-2.66x)=28.67

9x+31.44-10.64X=28.67

-1.64x=-2.77

x=1.69.

Therefore, each card costs 1.69. Plug that value back into the first equation again:

8(1.69)+3y=23.54

13.52+3y=23.54

3y=10.02

y=3.34.

So each box costs 3.34. 2 boxes then costs 6.68.