First student bought art, math and history books for $72. a+m+h=72
Second student bought art, chemistry and history books for $80. a+c+h=80
Third student bought chemistry, math and history books for $61. c+m+h=61
Fourth student bought art, math and chemistry books for $57. a+m+c=57
How much does each book cost?
How would you do this without guess and check?
You have 4 equations in 4 unknowns, so use elimination or substitution. For example,
a+0+h+m = 72
a+c+h+0 = 80
0+c+h+m = 61
a+c+0+m = 57
subtract #1 from #2 and #4
a+0+h+m = 72
0+c+0-m = 8
0+c+h+m = 61
0+c-h+0 = -15
subtract #2 from #3 and #4
a+0+h+m = 72
0+c+0-m = 8
0+0+h+2m = 53
0+0-h+m = -23
add #3 to #4
a+0+h+m = 72
0+c+0-m = 8
0+0+h+2m = 53
0+0+0+3m = 30
so, m=10
use that to fill in the other values, working your way up the rows.
subtracting 1st from 2nd ... c - m = 8
... c = m + 8
substitute into 3rd ... 2m + h = 53 (A)
substitute into 4th ... 2m + a = 49 (B)
subtracting ... h - a = 4
adding to 1st ... 2h + m = 76
... multiply by 2 ... 4h + 2m = 152
... subtract A ... 3h = 99 ... h = 33
substitute to find m , c , and a
To solve this problem without guess and check, we can use a method called "elimination."
First, let's assign variables to the cost of each book: a for art, m for math, c for chemistry, and h for history.
Now let's solve the system of equations using elimination:
From the first equation, we have a + m + h = 72.
From the second equation, we have a + c + h = 80.
From the third equation, we have c + m + h = 61.
From the fourth equation, we have a + m + c = 57.
To eliminate a variable, we can subtract equations. Let's start by eliminating the variable 'a' from the equations:
From the first equation, subtract the fourth equation: a + m + h - (a + m + c) = 72 - 57
This simplifies to: h - c = 15. (Equation 5)
Now let's eliminate the variable 'c' from the equations:
From the second equation, subtract the third equation: (a + c + h) - (c + m + h) = 80 - 61
This simplifies to: a - m = 19. (Equation 6)
Now we have a system of two equations (Equation 5 and Equation 6) with two variables (h and c). We can solve this system to find the values of h and c.
Multiply Equation 5 by -1: -h + c = -15. (Equation 7)
Add Equation 6 and Equation 7: (a - m) + (-h + c) = 19 - 15
This simplifies to: a - m - h + c = 4. (Equation 8)
Now, we can combine Equation 8 with the first equation (a + m + h = 72) to eliminate the 'a' variable:
Subtract Equation 8 from the first equation: (a + m + h) - (a - m - h + c) = 72 - 4
This simplifies to: 2h - c = 68. (Equation 9)
We now have two equations with two variables: Equation 7 (from the elimination of a) and Equation 9 (from the elimination of a and m). Let's solve this system to find the values of h and c.
Multiply Equation 7 by 2: -2h + 2c = -30. (Equation 10)
Add Equation 9 and Equation 10: (2h - c) + (-2h + 2c) = 68 - 30
This simplifies to: h + c = 38. (Equation 11)
Now that we have the value of h + c, we can substitute it back into the first equation (a + m + h = 72) to find the value of a + m:
a + m + 38 = 72
a + m = 72 - 38
a + m = 34. (Equation 12)
Now we have the value of a + m, and we can substitute it back into the fourth equation (a + m + c = 57) to find the value of c:
34 + c = 57
c = 57 - 34
c = 23.
Similarly, we can substitute the value of a + m into the second equation (a + c + h = 80) to find the value of h:
a + 23 + h = 80
a + h = 80 - 23
a + h = 57.
Finally, we can substitute the values of c and a + h into the first equation (a + m + h = 72) to find the value of m:
34 + m + 57 = 72
m = 72 - 34 - 57
m = -19.
Since the value of m is negative, it means there was an error in the given information or in the calculations. Please double-check the equations and values provided.