Use the system of equations you selected above to answer the question, "How many lbs of each ingredient did the store use?"
g+j+h=8
2g+3j+3h=20
g=4j
To solve the system of equations, we can substitute the value of g in terms of j into the first equation.
Since g=4j, we substitute this into the first equation:
4j+j+h=8
Simplifying, we get:
5j+h=8
Now, we can eliminate the variable h by subtracting the second equation from the third equation:
2g+3j+3h=20
-2g-4j = 0
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3j+3h-4j = 20-0
j+3h = 20
Now we have a system of two equations with two variables:
5j+h=8 ...(1)
j+3h=20 ...(2)
We can solve this system using substitution or elimination method.
Let's use substitution method:
From equation (2), we can rewrite j in terms of h:
j = 20-3h
Substituting this into equation (1), we get:
5(20-3h)+h=8
100-15h+h=8
100-14h=8
-14h=8-100
-14h=-92
h = (-92)/(-14)
h = 6.57 (rounded to two decimal places)
Now we can substitute the value of h back into equation (2) to find j:
j+3(6.57) = 20
j+19.71 = 20
j = 20-19.71
j = 0.29 (rounded to two decimal places)
Finally, we can substitute the values of j and h into equation (1) to find g:
5(0.29)+h=8
1.45+h=8
h = 8-1.45
h = 6.55 (rounded to two decimal places)
Therefore, the store used approximately g=0.29 lbs of one ingredient, j=6.55 lbs of another ingredient, and h=6.57 lbs of the third ingredient.
To solve the system of equations, let's substitute the value of g from the third equation into the first two equations.
Since g = 4j, we can replace g with 4j in the first and second equations, respectively.
So the new equations are:
4j + j + h = 8 (Equation 1)
2(4j) + 3j + 3h = 20 (Equation 2)
Simplifying Equation 1 gives us:
5j + h = 8 (Equation 3)
Simplifying Equation 2 gives us:
8j + 3h = 20 (Equation 4)
Now we have a system of two equations with two variables, j and h.
Next, let's solve this system of equations by using the method of substitution.
From Equation 3, we can isolate h:
h = 8 - 5j
Substitute this value of h into Equation 4:
8j + 3(8 - 5j) = 20
Simplify:
8j + 24 - 15j = 20
Combine like terms:
-7j + 24 = 20
Subtract 24 from both sides:
-7j = -4
Divide by -7:
j = -4 / -7
Simplify:
j = 4/7
Now that we have the value of j, we can substitute it back into Equation 3 to find h:
5(4/7) + h = 8
20/7 + h = 8
Subtract 20/7 from both sides:
h = 8 - 20/7
To simplify, find a common denominator for 7:
h = (56/7 - 20/7) / 7
h = 36/7 / 7
h = 36/49
Now we can find the value of g using Equation 1:
g + 4/7 + 36/49 = 8
We can find a common denominator for 7 and 49:
(49g + 4*7 + 36) / 49 = 8
(49g + 28 + 36) / 49 = 8
Combine like terms:
(49g + 64) / 49 = 8
Multiply both sides by 49:
49g + 64 = 392
Subtract 64 from both sides:
49g = 328
Divide by 49:
g = 328 / 49
Simplify the fraction:
g = 64/7
Therefore, the store used 64/7 lbs of ingredient g, 4/7 lbs of ingredient j, and 36/49 lbs of ingredient h.
To find the solution to this system of equations and determine the amount of each ingredient used, we can use the method of substitution.
First, we will use the equation g=4j and substitute it into the other two equations to eliminate the variable g.
Substituting g=4j into the first equation:
4j + j + h = 8
Combining like terms:
5j + h = 8 ----(Equation 1)
Substituting g=4j into the second equation:
2(4j) + 3j + 3h = 20
8j + 3j + 3h = 20
Combining like terms:
11j + 3h = 20 ----(Equation 2)
Now, we have a system of two equations in the variables j and h:
Equation 1: 5j + h = 8
Equation 2: 11j + 3h = 20
We can solve this system of equations using any method such as substitution, elimination, or matrices. Let's solve it by substitution.
From Equation 1, we can isolate h:
h = 8 - 5j
Now substitute this expression for h in Equation 2:
11j + 3(8 - 5j) = 20
11j + 24 - 15j = 20
Combining like terms:
-4j + 24 = 20
Subtract 24 from both sides:
-4j = -4
Divide both sides by -4:
j = 1
Now, substitute the value of j back into Equation 1 to get the value of h:
5(1) + h = 8
5 + h = 8
Subtract 5 from both sides:
h = 3
Finally, substitute j=1 and h=3 back into the equation g = 4j to get the value of g:
g = 4(1)
g = 4
Therefore, the solution to the system of equations and the amount of each ingredient used are:
g = 4 lbs
j = 1 lb
h = 3 lbs