Andres school orders some new supplies for the chemistry lab. The online store shows a pack of 10 test tubes costs $4 less than a set of nested beakers. in order to fully equip the lab, the school orders 12 sets of beakers and 8 packs of rest tubes. a. Write an equation that shows the cost of a pack of test tubes, t, in terms of the cost of a set of beakers. b.the school office receives a bill for the supplies in the amount of $348. Write an equation with t and b that describes this situation. d. solve equation for b. e. How much did the school pay for a set of beakers? for a pack of test tubes?

10t = b - 4

12b+8t = $348

This is a system of equations. I’ll be solving through substitution.

In the first equation. solving for b (the easier variable to isolate) gives you:

b = 10t + 4

Substitute this into the second equation:

12(10t+4) +8t = 348
120t+48+8t = 348
128t = 300

t = 2.34375 —> round it to the nearest cent to get 2.34 dollars

b = 10t+4
b = 10(2.34)+4
b = 27.4 dollars

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MrRoyal MrRoyal · Answer 2 · 2021-11-02T15:46:14+01:00

The question is an illustration of system of equations.

The school paid $27.4 for a set of beakers and $2.34 for a pack of test tubes.

Represent the number of beakers with b and the number of test tubes with t.

(a) Equation that expresses t in terms of b

The statements from the question can be represented as:

[tex]\mathbf{10t = b - 4}[/tex]

(b) Equation that expresses t and b

The second statement from the question can be represented as:

[tex]\mathbf{12b + 8t = 348}[/tex]

So, we have:

[tex]\mathbf{10t = b - 4}[/tex]

[tex]\mathbf{12b + 8t = 348}[/tex]

(c) Solve for b and t

Make b the subject in [tex]\mathbf{10t = b - 4}[/tex]

[tex]\mathbf{b = 10t + 4}[/tex]

Substitute [tex]\mathbf{b = 10t + 4}[/tex] in [tex]\mathbf{12b + 8t = 348}[/tex]

[tex]\mathbf{12(10t + 4) + 8t = 348}[/tex]

[tex]\mathbf{120t + 48 + 8t = 348}[/tex]

Collect like terms

[tex]\mathbf{120t + 8t = 348 - 48}[/tex]

[tex]\mathbf{128t = 300}[/tex]

Divide both sides by 128

[tex]\mathbf{t = 2.34}[/tex]

So, we have:

[tex]\mathbf{b = 10t + 4}[/tex]

Substitute [tex]\mathbf{t = 2.34}[/tex]

[tex]\mathbf{b = 2.34 \times 10 + 4}[/tex]

[tex]\mathbf{b = 27.4}[/tex]

Hence, the school paid $27.4 for a set of beakers and $2.34 for a pack of test tubes.

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