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Answer:

[tex]\boxed {\boxed {\sf 3 \ pounds \ of \$4 \ candy}}[/tex]

[tex]\boxed {\boxed {\sf 2.5 \ pounds \ of \$3.50 \ candy}}[/tex]

Step-by-step explanation:

Let's create a system of equation. First, define the variables. One type of candy is x and the other is y.

Now, create 2 equations, one for the weight and one for the price.

[tex]x+y=5.5[/tex]

[tex]4x+3.5y=20.75[/tex]

The first equation represents the mixture of both types of candy that weighs 5.5 pounds. The second represents the prices of the candy and the mixture.

Next. solve for the variables using substitution. Isolate a variable in the first equation. Let's isolate y by subtracting x.

[tex]x-x+y=5.5-x\\y= 5.5-x[/tex]

Now, we know what y is equal to, so we can substitute it into the second equation.

[tex]4x+ 3.5 (5.5-x)= 20.75[/tex]

Distribute the 3.5. Multiply each term by 3.5

[tex]4x+(3.5*5.5)+(3.5*-x)[/tex]

[tex]4x+19.25-3.5x= 20.75[/tex]

Combine like terms on the right side.

[tex]0.5x+19.25=20.75[/tex]

Subtract 19.25 from both sides of the equation to isolate the variable.

[tex]0.5x+19.25-19.25=20.75-19.25[/tex]

[tex]0.5x= 1.5[/tex]

Divide both sides by 0.5.

[tex]\frac {0.5x}{0.5}= \frac{1.5}{0.5}\\x=3[/tex]

Now we can substitute x back into the first equation and solve for y.

[tex]x+y=5.5\\3+y=5.5[/tex]

Subtract 3 from both sides to isolate the variable.

[tex]3-3+y=5.5-3\\y=2.5[/tex]

There was 3 pounds of the $4 candy and 2.5 pounds of the $3.50 candy.

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