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PSAT/NMSQT Math / LEVEL 1 · DIFFICULTY 3/5

Algebra: Models & Constraints

Represent relationships with equations, lines, systems and inequalities.

4 stages · 32 practice problems · two 8-question assessment forms

STAGE PSAT 1.1 · 8 PRACTICE PROBLEMS · 4 NEW-CONTEXT APPLICATIONS

Linear Equations & Function Models

Goal: Translate a situation or table into a linear rule, solve with variables on both sides, and recognize equations with no or infinitely many solutions.

Before you begin: Read the problem carefully and use the linked foundation lessons when a term or operation is unfamiliar.

Parent line: y=x\displaystyle y = x. Change slope and intercept to see how a linear rule changes.

AN EQUATION IS A RULE FOR MAKING POINTS

Change the rule. Watch the graph.

Pick an x value. The rule tells you its y value. A point (x, y) shows both numbers together. Drag across the graph or use the point slider. This is an adjustable learning example; use the coefficients in your own practice question when solving it.

Parent equation: y=x\displaystyle y = x

Start with y=x\displaystyle y = x. Multiply each y value by 2, then shift up 1. A negative multiplier reflects across the x-axis.

Transform the parent graph
0%

Teal moving graph: y=1 x+(0)\displaystyle y = 1\, x + \left(0\right)

Blue: y=2 x+1\displaystyle y = 2\, x + 1

y = 2x + 1; horizontal x-axis and vertical y-axis-10-10-8-8-6-6-4-4-2-2224466881010xy0

Moving point: (-4, -4)

Teal shows the current transformation. Blue is the target. At 100%\displaystyle 100\% they coincide; intermediate graphs are learning frames, not additional solutions.

Dashed gray: parent equation. Blue: current equation. View: −10 to 10 on both axes. The graph continues beyond this window. Purple marks your selected point.

Slope a: 2
Vertical intercept b: 1
Choose the point’s x: 1
Point (1, 3)
2×(1)+(1)=3\displaystyle 2 \times \left(1\right) + \left(1\right) = 3
What does moving one step right do?

Each step of 1 to the right changes y by 2. The line rises. Setting x=0\displaystyle x = 0 gives y=1\displaystyle y = 1, where the line meets the vertical axis.

Three points made by the blue rule
xSubstitute into the ruley
-12×(−1)+(1)\displaystyle 2 \times \left(-1\right) + \left(1\right)-1
02×(0)+(1)\displaystyle 2 \times \left(0\right) + \left(1\right)1
12×(1)+(1)\displaystyle 2 \times \left(1\right) + \left(1\right)3

Make sense of the idea

A linear rule has a constant change. In f(x) = mx + b, x is the input, f(x) is the output, m is change per input unit, and b is the output at zero.

Choose a method

Distribute across every term in brackets, then collect variable terms on one side and constants on the other. Do the same operation to both sides.

Check and interpret

If the variables cancel, read what remains: 0=0\displaystyle 0 = 0 is true for every input; 0=5\displaystyle 0 = 5 is impossible. A context may further restrict the inputs to nonnegative values or whole numbers.

WORKED EXAMPLE 1

Solve 5 (x+2)=4 x+17\displaystyle 5\, \left(x + 2\right) = 4\, x + 17.

  1. Distribute to get 5 x+10=4 x+17\displaystyle 5\, x + 10 = 4\, x + 17.
  2. Subtract 4x and 10:x=7\displaystyle 10 : x = 7. Substitute into both original costs to check.

7

WORKED EXAMPLE 2

Two balance records require 6 x+4=6 x+7\displaystyle 6\, x + 4 = 6\, x + 7. How many real solutions does this equation have?

  1. Subtract 6x to obtain 4=7\displaystyle 4 = 7.
  2. This is false independently of x. There are 0 solutions.

0

Common pitfalls

Possible mix-up: An equation always has one answer.

An identity has infinitely many real solutions; a contradiction has none.

Possible mix-up: The intercept is the rate.

The intercept is the starting value. The slope is the change per unit.

Explain it to yourself

In linear equations & function models, what information determines your method, and how can you check whether your result fits the original conditions?

Preview the eight practice prompts
  1. Problem 1

    Solve 3 (x+2)=2 x+11\displaystyle 3\, \left(x + 2\right) = 2\, x + 11.

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  2. Problem 2

    A linear function has f (0)=7\displaystyle f\, \left(0\right) = 7 and f (2)=11\displaystyle f\, \left(2\right) = 11. Find f(3).

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  3. Problem 3

    Consider 2 (x+3)=2 x+6\displaystyle 2\, \left(x + 3\right) = 2\, x + 6. How many real solutions are there? Enter 0 for none, 1 for one, or 2 for infinitely many.

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  4. Problem 4

    Consider 3 x+1=3 x+4\displaystyle 3\, x + 1 = 3\, x + 4. How many real solutions does this equation have?

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  5. Problem 5New context

    Two delivery plans cost 4 (x+2)\displaystyle 4\, \left(x + 2\right) dollars and 3 x+14\displaystyle 3\, x + 14 dollars for x trips. At how many trips do they cost the same?

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  6. Problem 6New context

    A tank contains 8 liters initially. A pump adds 3 liters per minute. If V(t) is the volume after t minutes, find V(3).

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  7. Problem 7New context

    Two club fee formulas give 3 (x+3)=3 x+9\displaystyle 3\, \left(x + 3\right) = 3\, x + 9. How many real solutions are there? Enter 0 for none, 1 for one, or 2 for infinitely many.

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  8. Problem 8New context

    Two balance records require 4 x+2=4 x+5\displaystyle 4\, x + 2 = 4\, x + 5. How many real solutions does this equation have?

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Open stage PSAT 1.1 in the student workspace →

STAGE PSAT 1.2 · 8 PRACTICE PROBLEMS · 4 NEW-CONTEXT APPLICATIONS

Lines: Forms, Slopes & Parameters

Goal: Connect standard, slope-intercept and point-slope forms; interpret slope and find parallel or perpendicular lines.

Before you begin: Read the problem carefully and use the linked foundation lessons when a term or operation is unfamiliar.

Parent line: y=x\displaystyle y = x. Change slope and intercept to see how a linear rule changes.

AN EQUATION IS A RULE FOR MAKING POINTS

Change the rule. Watch the graph.

Pick an x value. The rule tells you its y value. A point (x, y) shows both numbers together. Drag across the graph or use the point slider. This is an adjustable learning example; use the coefficients in your own practice question when solving it.

Parent equation: y=x\displaystyle y = x

Start with y=x\displaystyle y = x. Multiply each y value by 2, then shift up 1. A negative multiplier reflects across the x-axis.

Transform the parent graph
0%

Teal moving graph: y=1 x+(0)\displaystyle y = 1\, x + \left(0\right)

Blue: y=2 x+1\displaystyle y = 2\, x + 1

y = 2x + 1; horizontal x-axis and vertical y-axis-10-10-8-8-6-6-4-4-2-2224466881010xy0

Moving point: (-4, -4)

Teal shows the current transformation. Blue is the target. At 100%\displaystyle 100\% they coincide; intermediate graphs are learning frames, not additional solutions.

Dashed gray: parent equation. Blue: current equation. View: −10 to 10 on both axes. The graph continues beyond this window. Purple marks your selected point.

Slope a: 2
Vertical intercept b: 1
Choose the point’s x: 1
Point (1, 3)
2×(1)+(1)=3\displaystyle 2 \times \left(1\right) + \left(1\right) = 3
What does moving one step right do?

Each step of 1 to the right changes y by 2. The line rises. Setting x=0\displaystyle x = 0 gives y=1\displaystyle y = 1, where the line meets the vertical axis.

Three points made by the blue rule
xSubstitute into the ruley
-12×(−1)+(1)\displaystyle 2 \times \left(-1\right) + \left(1\right)-1
02×(0)+(1)\displaystyle 2 \times \left(0\right) + \left(1\right)1
12×(1)+(1)\displaystyle 2 \times \left(1\right) + \left(1\right)3

Make sense of the idea

Parent equation: y=x\displaystyle y = x. A line y = mx + b changes the slope and shifts the intercept. To read slope from Ax + By = C, isolate y first.

Choose a method

Point-slope form y−y1=m (x−x1)\displaystyle y - y_{1} = m\, \left(x - x_{1}\right) records one point and a slope. Substitute the point to verify that it lies on the line.

Check and interpret

Parallel nonvertical lines have equal slopes. Perpendicular nonvertical lines have slopes whose product is −1. Horizontal and vertical lines are a separate perpendicular pair.

WORKED EXAMPLE 1

For the line 8 x+2 y=6\displaystyle 8\, x + 2\, y = 6, what is the slope?

  1. 2 y=−8 x+6\displaystyle 2\, y = -8\, x + 6.
  2. Thus y=−4 x+3\displaystyle y = -4\, x + 3, whose slope is −4.

-4

WORKED EXAMPLE 2

A sensor calibration table lists x=1\displaystyle x = 1, 3, 5 and y=11\displaystyle y = 11, 21, 31, respectively. Find the value of y when x=0\displaystyle x = 0.

  1. The slope is 21−113−1=5\displaystyle \frac{21 - 11}{3 - 1} = 5.
  2. The intercept is 11−5=6\displaystyle 11 - 5 = 6.

6

Common pitfalls

Possible mix-up: The coefficient of x in standard form is the slope.

Solve for y; the slope is −AB\displaystyle \frac{-A}{B} when B is nonzero.

Possible mix-up: A perpendicular slope only changes the sign.

It is the negative reciprocal, so the product of the two slopes is −1.

Explain it to yourself

In lines: forms, slopes & parameters, what information determines your method, and how can you check whether your result fits the original conditions?

Preview the eight practice prompts
  1. Problem 1

    For the line 4 x+2 y=2\displaystyle 4\, x + 2\, y = 2, what is the slope?

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  2. Problem 2

    A line through (1, 8) is parallel to y=2 x+1\displaystyle y = 2\, x + 1. Find its y-intercept.

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  3. Problem 3

    A perpendicular line to y=2 x+4\displaystyle y = 2\, x + 4. What slope must the other line have?

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  4. Problem 4

    A linear table lists x=1\displaystyle x = 1, 3, 5 and y=5\displaystyle y = 5, 9, 13, respectively. Find the value of y when x=0\displaystyle x = 0.

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  5. Problem 5New context

    A straight path is modeled by 6 x+2 y=4\displaystyle 6\, x + 2\, y = 4, what is the slope?

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  6. Problem 6New context

    A parallel route through (2, 11) is parallel to y=3 x+1\displaystyle y = 3\, x + 1. Find its y-intercept.

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  7. Problem 7New context

    Two straight beams are perpendicular. One beam lies on y=3 x+5\displaystyle y = 3\, x + 5. What is the slope of the other beam?

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  8. Problem 8New context

    A sensor calibration table lists x=1\displaystyle x = 1, 3, 5 and y=7\displaystyle y = 7, 13, 19, respectively. Find the value of y when x=0\displaystyle x = 0.

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Open stage PSAT 1.2 in the student workspace →

STAGE PSAT 1.3 · 8 PRACTICE PROBLEMS · 4 NEW-CONTEXT APPLICATIONS

Linear Systems: Solutions & Constraints

Goal: Solve contextual systems and use proportional coefficients to distinguish one, zero and infinitely many solutions.

Before you begin: Read the problem carefully and use the linked foundation lessons when a term or operation is unfamiliar.

Parent line: y=x\displaystyle y = x. Change slope and intercept to see how a linear rule changes.

AN EQUATION IS A RULE FOR MAKING POINTS

Change the rule. Watch the graph.

Pick an x value. The rule tells you its y value. A point (x, y) shows both numbers together. Drag across the graph or use the point slider. This is an adjustable learning example; use the coefficients in your own practice question when solving it.

Parent equation: y=x\displaystyle y = x

Start with y=x\displaystyle y = x. Multiply each y value by 2, then shift up 1. A negative multiplier reflects across the x-axis.

The orange line also starts from y=x\displaystyle y = x: reflect to y=−x\displaystyle y = -x, then move up 4.

Transform the parent graph
0%

Teal moving graph: y=1 x+(0)\displaystyle y = 1\, x + \left(0\right)

Blue: y=2 x+1\displaystyle y = 2\, x + 1
Orange: y=−x+4\displaystyle y = -x + 4

y = 2x + 1; horizontal x-axis and vertical y-axis-10-10-8-8-6-6-4-4-2-2224466881010xy0

Moving point: (-4, -4)

Teal shows the current transformation. Blue is the target. At 100%\displaystyle 100\% they coincide; intermediate graphs are learning frames, not additional solutions.

Dashed gray: parent equation. Blue: current equation. View: −10 to 10 on both axes. The graph continues beyond this window. Purple marks your selected point.

Slope a: 2
Vertical intercept b: 1
Choose the point’s x: 1
Point (1, 3)
2×(1)+(1)=3\displaystyle 2 \times \left(1\right) + \left(1\right) = 3
What does moving one step right do?

Each step of 1 to the right changes y by 2. The line rises. Setting x=0\displaystyle x = 0 gives y=1\displaystyle y = 1, where the line meets the vertical axis.

Three points made by the blue rule
xSubstitute into the ruley
-12×(−1)+(1)\displaystyle 2 \times \left(-1\right) + \left(1\right)-1
02×(0)+(1)\displaystyle 2 \times \left(0\right) + \left(1\right)1
12×(1)+(1)\displaystyle 2 \times \left(1\right) + \left(1\right)3

Make sense of the idea

A solution to a system must satisfy both equations at once. Graphically it is a common point. A crossing gives one point; distinct parallel lines give none; the same line gives infinitely many.

Choose a method

Elimination combines equations to remove one variable. Multiplying an equation means multiplying every term, including the constant.

Check and interpret

For a parameter problem compare the ratios of the coefficients. Proportional x- and y-coefficients need proportional constants too for the equations to describe the same line.

WORKED EXAMPLE 1

Solve x+y=12\displaystyle x + y = 12 and 5 x+2 y=39\displaystyle 5\, x + 2\, y = 39. Find x.

  1. Subtracting gives 3 x=15\displaystyle 3\, x = 15.
  2. So x=5\displaystyle x = 5; y=7\displaystyle y = 7. Both equations check.

5

WORKED EXAMPLE 2

Two measurement relationships uses y=5 x+1\displaystyle y = 5\, x + 1 and y=6 x+6\displaystyle y = 6\, x + 6. Find the x-coordinate of the intersection.

  1. 5 x+1=6 x+6\displaystyle 5\, x + 1 = 6\, x + 6.
  2. Subtract 5x and 6:x=−5\displaystyle 6 : x = -5. Different slopes ensure exactly one intersection.

-5

Common pitfalls

Possible mix-up: Matching slopes means the lines are identical.

The intercepts must also match. Distinct parallel lines never meet.

Possible mix-up: Any point on either line solves the system.

A common solution must satisfy both equations.

Explain it to yourself

In linear systems: solutions & constraints, what information determines your method, and how can you check whether your result fits the original conditions?

Preview the eight practice prompts
  1. Problem 1

    Solve x+y=8\displaystyle x + y = 8 and 5 x+2 y=25\displaystyle 5\, x + 2\, y = 25. Find x.

    Practise this exact problem →
  2. Problem 2

    The system is x+2 y=4\displaystyle x + 2\, y = 4 and 3 x+6 y=13\displaystyle 3\, x + 6\, y = 13. How many solutions are there?

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  3. Problem 3

    Two equations are x+2 y=5\displaystyle x + 2\, y = 5 and 4 x+8 y=c\displaystyle 4\, x + 8\, y = c. What value of c makes them the same line?

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  4. Problem 4

    A system uses y=2 x+1\displaystyle y = 2\, x + 1 and y=3 x+3\displaystyle y = 3\, x + 3. Find the x-coordinate of the intersection.

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  5. Problem 5New context

    A club sells 10 tickets. Adult tickets cost $5 and child tickets cost $2. Revenue is $32. How many adult tickets were sold?

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  6. Problem 6New context

    Two model constraints are x+2 y=5\displaystyle x + 2\, y = 5 and 3 x+6 y=16\displaystyle 3\, x + 6\, y = 16. How many solutions are there?

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  7. Problem 7New context

    Two line models are x+2 y=6\displaystyle x + 2\, y = 6 and 4 x+8 y=c\displaystyle 4\, x + 8\, y = c. What value of c makes them the same line?

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  8. Problem 8New context

    Two measurement relationships uses y=3 x+1\displaystyle y = 3\, x + 1 and y=4 x+4\displaystyle y = 4\, x + 4. Find the x-coordinate of the intersection.

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Open stage PSAT 1.3 in the student workspace →

STAGE PSAT 1.4 · 8 PRACTICE PROBLEMS · 4 NEW-CONTEXT APPLICATIONS

Graphing Inequalities & Feasible Regions

Goal: Choose a solid or dashed boundary, test points in half-planes, and solve restrictions by intersecting shaded regions.

Before you begin: Read the problem carefully and use the linked foundation lessons when a term or operation is unfamiliar.

Find the overlap. Test your point.

Parent line: y=x\displaystyle y = x. We move it up by b, then keep points below it and above y=2\displaystyle y = 2. The blue overlap is the feasible region in the displayed window.

001122334455667788xy

(2, 4) satisfies both conditions. Compare 4 with the maximum 4, and also check 4≥2\displaystyle 4 \ge 2.

Try placing your point exactly on the sloping boundary, then switch solid to dashed. Equality is the only difference.

Make sense of the idea

Replace the inequality symbol with = to draw its boundary. Use a solid boundary for ≤ or ≥ because equality is allowed. Use a dashed boundary for < or > because the boundary itself is excluded.

Choose a method

For y > mx + b shade above the line; for y < mx + b shade below. You can always check a point by substitution. When dividing an inequality by a negative number, reverse its direction.

Check and interpret

A system requires every inequality. Shade only the overlap. In a budget or capacity problem also include nonnegative quantities and, when counting objects, whole-number restrictions.

WORKED EXAMPLE 1

Consider y<2 x+5\displaystyle y < 2\, x + 5. Should its boundary be solid or dashed? Enter 1 for solid or 2 for dashed.

  1. The boundary is y=2 x+5\displaystyle y = 2\, x + 5.
  2. Equality is excluded, so use a dashed line and enter 2.

2

WORKED EXAMPLE 2

A sensor value must meet −2 x≥12\displaystyle -2\, x \ge 12. What is the greatest allowed x?

  1. Dividing gives x≤−6\displaystyle x \le -6.
  2. The endpoint is included, so the greatest allowed x is -6.

-6

Common pitfalls

Possible mix-up: A point on a dashed line is allowed.

A strict inequality excludes equality.

Possible mix-up: Meeting one condition is enough.

A feasible point satisfies every condition simultaneously.

Explain it to yourself

In graphing inequalities & feasible regions, what information determines your method, and how can you check whether your result fits the original conditions?

Preview the eight practice prompts
  1. Problem 1

    Consider y<2 x+3\displaystyle y < 2\, x + 3. Should its boundary be solid or dashed? Enter 1 for solid or 2 for dashed.

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  2. Problem 2

    Test the point (2, 6) against y≥2 x+3\displaystyle y \ge 2\, x + 3. Is it feasible? Enter 1 for yes or 0 for no.

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  3. Problem 3

    For x≥0\displaystyle x \ge 0, y≥3\displaystyle y \ge 3 and 2 x+y≤18\displaystyle 2\, x + y \le 18, find the greatest y when x=3\displaystyle x = 3.

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  4. Problem 4

    Solve −2 x≥6\displaystyle -2\, x \ge 6. What is the greatest allowed x?

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  5. Problem 5New context

    A safety rule requires y<2 x+4\displaystyle y < 2\, x + 4. Should its boundary be solid or dashed? Enter 1 for solid or 2 for dashed.

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  6. Problem 6New context

    A proposed production plan is (3, 9) and must satisfy y≥2 x+4\displaystyle y \ge 2\, x + 4. Is it feasible? Enter 1 for yes or 0 for no.

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  7. Problem 7New context

    A workshop makes x small kits and y large kits. Its budget requires 2 x+y≤20\displaystyle 2\, x + y \le 20, and y≥3\displaystyle y \ge 3. With x=4\displaystyle x = 4, what is the greatest whole number of large kits possible?

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  8. Problem 8New context

    A sensor value must meet −2 x≥8\displaystyle -2\, x \ge 8. What is the greatest allowed x?

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Open stage PSAT 1.4 in the student workspace →
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