Solve a practical problem with simultaneous equations lesson | Year 10 Mathematics (Linear equations and inequalities, Simultaneous equations by substitution)

Use and check a contextual pair

This topic combines setting up a model with solving and interpreting it. A final value must fit both conditions and make sense for the quantity being counted or measured

Define the quantities Distinguish an item count from a unit price. Keep the same variable definitions in both conditions
Choose a short route Use substitution when one quantity is easy to isolate. Use elimination when a pair of coefficients already matches or can be matched with small multipliers
Keep exact money values Use exact decimal amounts or work in cents. Do not round during the solution
Check and interpret Check both original conditions. Report the requested quantity with its unit, and reject negative or fractional item counts when the context requires whole items
State the assumption A two-price sales model assumes every item in a category has the same price and the recorded total has no extra charge or discount

Worked example

A maker sells small storage tubs for $5 each and large storage tubs for $9 each. A sale of 32 tubs brings in $224. Find the number of each size sold

\begin{gathered}s+l=32\\5s+9l=224\end{gathered}

$\displaystyle s=32-l$Express the small-tub count in terms of the large-tub count
$\displaystyle 5(32-l)+9l=224$Substitute into the money equation
$\displaystyle 160-5l+9l=224$Expand the bracket
$\displaystyle 4l=64$Collect terms and subtract 160.
$\displaystyle l=16$Divide by 4.
$\displaystyle s=32-16=16$Find the other count
$\displaystyle 16+16=32$Check the original count
$\displaystyle 5(16)+9(16)=224$Check the original takings. Sixteen tubs of each size were sold

Common mistakes

Checking only the item total

✕$\displaystyle \begin{aligned}p&=10\\b&=10\\p+b&=20\end{aligned}$✓$\displaystyle \begin{aligned}7(10)+2(10)&=90\\90&\ne85\end{aligned}$

For the pouch-and-badge model, the proposed pair satisfies the count condition but fails the takings condition

Reporting the other quantity

✕$\displaystyle \text{Lantern price}=\$4$✓$\displaystyle \text{Lantern price}=\$7$

Two balloons and three lanterns cost $29; five balloons and three lanterns cost $41. Subtracting the orders shows that three balloons cost $12, so one balloon costs $4. The first order then shows that three lanterns cost $21, so one lantern costs $7

Scaling only the left side

✕$\displaystyle 5s+5l=32$✓$\displaystyle 5s+5l=160$

In the storage-tub example, multiplying the count equation by five requires multiplying its right side as well

Difficulty levels

introFind two unit prices from two short purchase records

standardFind the requested item count from a count total and takings

examChoose the requested count from a short practical problem without setup prompts

Questions about this topic

Must I use elimination?+

No. Use substitution, elimination or another valid method. Check the final pair against both original conditions

Why check the context after solving?+

The equations may give values that do not fit a physical situation. A count of whole items cannot be negative or fractional

What if the two prices are equal?+

For a count-and-value model, the money total then repeats the count information or contradicts it. It cannot determine the split. These practice questions use different prices

Can I round a fractional item count?+

No. If the context requires whole items, a fractional solution means the stated model or data needs checking. Rounding generally breaks one or both equations

Can I use a calculator?+

A calculator is allowed unless a question is marked no calculator

Which year level and chapter is this?+

Year 10 Mathematics, chapter 8: Linear equations and inequalities

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Solve a practical problem with simultaneous equations

Contents

Solve, then return to the situation

Checks to finish