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Clint is making a 10-lb bag of trail mix for his upcoming backpacking trip. Thechocolates cost $3.00 per pound and mixed nuts cost $6.00 per pound and Clint has abudget of $5.10 per pound of trail mix. Using the variables c and n to represent thenumber of pounds of chocolate and the number of pounds of nuts he should userespectively, determine a system of equations that describes the situation.Enter the equations below separated by a comma.How many pounds of chocolate should he use?How many pounds of mixed nuts should he use? Pls see the picture

Clint Is Making A 10lb Bag Of Trail Mix For His Upcoming Backpacking Trip Thechocolates Cost 300 Per Pound And Mixed Nuts Cost 600 Per Pound And Clint Has Abudg class=

Sagot :

Since c represents the number of pounds of chocolates

Since n represents the number of pounds of nuts

Since Clint is making 10 pounds of them, then

[tex]c+n=10\rightarrow(1)[/tex]

Since the cost of 1 pound of chocolates is $3.00

Since the cost of 1 pound of nuts is $6.00

Since the Clint budget is $5.1 per pound, then

[tex]10\times5.1-\text{ \$51}[/tex]

Multiply c by 3 and n by 6, then add the products and equate the sum by 51

[tex]\begin{gathered} 3(c)+6(n)=5.1 \\ 3c+6n=51\rightarrow(2) \end{gathered}[/tex]

The system of equations is

c + n = 10

3c + 6n = 51

Let us solve them

Multiply equation (1) by -3 to make the coefficient of c equal in values and different in signs

[tex]\begin{gathered} -3(c)-3(n)=-3(10) \\ -3c-3n=-30\rightarrow(3) \end{gathered}[/tex]

Add equations (2) and (3)

[tex]\begin{gathered} (3c-3c)+(6n-3n)=(51-30) \\ 0+3n=21 \\ 3n=21 \end{gathered}[/tex]

Divide both sides by 3 to find n

[tex]\begin{gathered} \frac{3n}{3}=\frac{21}{3} \\ n=7 \end{gathered}[/tex]

Substitute n by 7 in equation (1)

[tex]c+7=10[/tex]

Subtract 7 from both sides

[tex]\begin{gathered} c+7-7=10-7 \\ c=3 \end{gathered}[/tex]

He should use 3 pounds of chocolate

He should use 7 pounds of nuts

If his budget is $51

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