Modeling with Systems

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Use the original graphs, tables, and diagrams from released STAAR questions to practice reading labels, values, and relationships before answering the items below.

Archived released STAAR Algebra II EOC · 2013 · Item 3 · 1 point

Worked example

A school sells adult tickets for 8 dollars and student tickets for 5 dollars. It can sell at most 300 tickets and needs at least 1,800 dollars. Write the system.

1. Let a be the number of adult tickets and s the number of student tickets.
2. The ticket limit gives $a+s\le 300$.
3. The revenue goal gives $8a+5s\ge 1800$.
4. Counts cannot be negative: $a\ge 0$ and $s\ge 0$.

$a+s\le 300$, $8a+5s\ge 1800$, $a\ge 0$, $s\ge 0$

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TEKSForge practice · unscored

1. 1. A rectangular garden uses a house wall as one length and fencing on the other three sides. There are at most 50 feet of fencing. Write the fencing constraint.

   Show answer and reasoning

   $l+2w\le 50$

   The fence covers one length and two widths. “At most” includes 50, so the symbol is ≤.

2. 2. A bakery makes muffins, m, taking 2 minutes each, and cookies, c, taking 1 minute each. It has at most 120 minutes and needs at least 30 items. Write the system.

   Show answer and reasoning

   $2m+c\le 120$ and $m+c\ge 30$

   One inequality limits time and the other sets the minimum number of items. Both variables must also be nonnegative whole numbers.

Released STAAR practice

Texas Education Agency · archived 2013 released EOC items · scored with TEA's answer key; step-by-step explanations written by TEKSForge because this release has no rationales. Original graphs, tables, and diagrams are shown where the released item includes one.

Archived released STAAR Algebra II EOC · 2013 · Item 3 · 1 point

The rectangular patio shown in the diagram is enclosed on three sides by a fence. The area is no more than 400 square feet, and the total length of the three fenced sides is no more than 60 feet. Which system of inequalities can be used to find values of $l$, the patio's length, and $w$, the patio's width, that satisfy these conditions?

1. $l + 2w \le 60$ and $lw \le 400$
2. $2l + 2w \le 60$ and $lw \le 400$
3. $l + 2w \ge 60$ and $lw \ge 400$
4. $2l + 2w \ge 60$ and $lw \ge 400$

Show STAAR answer and explanation

$l + 2w \le 60$ and $lw \le 400$

Only one length and two widths are fenced, so the fence constraint is $l+2w\le60$. Rectangular area is $lw$, and ‘no more than 400’ means $lw\le400$.

$2l + 2w \le 60$ and $lw \le 400$: This counts all four sides of the rectangle, but the house replaces one length of fence.

$l + 2w \ge 60$ and $lw \ge 400$: ‘No more than’ requires ≤, not ≥, for both the fence length and area.

$2l + 2w \ge 60$ and $lw \ge 400$: This counts four fenced sides and reverses both upper-bound inequalities.

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Archived released STAAR Algebra II EOC · 2013 · Item 34 · 1 point

The hypotenuse of a right triangle is 17 inches long. A leg of this triangle, $y$, is 1 inch less than twice the length of the other leg, $x$. Which system of equations can be used to determine the lengths of the 2 legs of this right triangle in inches?

1. $x^2 + y^2 = 289$ and $y = 2x - 1$
2. $x^2 + y^2 = 289$ and $2x + y = 1$
3. $x + y = 17$ and $y = 2x - 1$
4. $x + y = 17$ and $2x + y = 1$

Show STAAR answer and explanation

$x^2 + y^2 = 289$ and $y = 2x - 1$

The Pythagorean theorem gives $x^2+y^2=17^2=289$. “1 inch less than twice x” is $y=2x-1$. (The legs are 8 and 15 inches.)

$x^2 + y^2 = 289$ and $2x + y = 1$: $2x+y=1$ says a sum equals 1, not that y is 1 less than twice x.

$x + y = 17$ and $y = 2x - 1$: The legs of a right triangle do not add to the hypotenuse; use the Pythagorean theorem.

$x + y = 17$ and $2x + y = 1$: Both equations misstate the relationships in the problem.

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