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Digital SAT Math: systems of Non linear equations in two variables-Practice Questions

Digital SAT Math: systems of Non linear equations in two variables-Practice Questions - New Syllabus

DSAT MAth Practice questions – all topics

  • Advanced Math Weightage: 35% Questions: 13-15
    • Equivalent expressions
    • Nonlinear equations in one variable and systems of equations in two variables
    • Nonlinear functions

DSAT MAth and English  – full syllabus practice tests

Question Easy

\( x – y = 1 \)

\( x + y = x^2 – 3 \)

Which ordered pair is a solution to the system of equations above?

A. \( (1 + \sqrt{3}, \sqrt{3}) \)

B. \( (\sqrt{3}, -\sqrt{3}) \)

C. \( (1 + \sqrt{5}, \sqrt{5}) \)

D. \( (\sqrt{5}, -1 + \sqrt{5}) \)

▶️ Answer/Explanation
Solution

Ans: A

Solve the system:

Add equations: \( (x – y) + (x + y) = 1 + (x^2 – 3) \).

\( 2x = x^2 – 2 \).

\( x^2 – 2x – 2 = 0 \).

Discriminant: \( (-2)^2 – 4(1)(-2) = 4 + 8 = 12 \).

\( x = \frac{2 \pm \sqrt{12}}{2} = \frac{2 \pm 2\sqrt{3}}{2} = 1 \pm \sqrt{3} \).

For \( x = 1 + \sqrt{3} \):

First equation: \( y = x – 1 = (1 + \sqrt{3}) – 1 = \sqrt{3} \).

Second equation: \( x + y = (1 + \sqrt{3}) + \sqrt{3} = 1 + 2\sqrt{3} \).

Right side: \( x^2 – 3 = (1 + \sqrt{3})^2 – 3 = (1 + 2\sqrt{3} + 3) – 3 = 1 + 2\sqrt{3} \).

Both sides match, so \( (1 + \sqrt{3}, \sqrt{3}) \) is a solution.

Choice B: \( \sqrt{3} – (-\sqrt{3}) = 2\sqrt{3} \neq 1 \).

Choice C: \( (1 + \sqrt{5}) – \sqrt{5} = 1 \), but second equation fails.

Choice D: \( \sqrt{5} – (-1 + \sqrt{5}) = 1 \), but second equation fails.

Question Easy

\( P = \frac{W}{t} \)

The power \( P \) produced by a machine is represented by the equation above, where \( W \) is the work performed during an amount of time \( t \). Which of the following correctly expresses \( W \) in terms of \( P \) and \( t \)?

A. \( W = Pt \)

B. \( W = \frac{P}{t} \)

C. \( W = \frac{t}{P} \)

D. \( W = P + t \)

▶️ Answer/Explanation
Solution

Ans: A

Start with: \( P = \frac{W}{t} \).

Multiply both sides by \( t \): \( Pt = W \).

Rewrite: \( W = Pt \).

Choice B: Incorrect, divides \( P \) by \( t \).

Choice C: Incorrect, inverts the relationship.

Choice D: Incorrect, adds \( P \) and \( t \).

Question Easy

\( E = 18,000 – 2,000t \)

\( V = 18,000 \cdot (0.85^t) \)

The given equations are two different models that can be used to find the value, in dollars, of a particular car \( t \) years after it was purchased. Which of the following statements correctly compares the values of \( E \) and \( V \) for \( 0 < t < 9 \)?

A. \( E \) is always less than \( V \).

B. \( E \) is always greater than \( V \).

C. \( E \) is initially greater than \( V \) but eventually becomes less than \( V \).

D. \( E \) is initially less than \( V \) but eventually becomes greater than \( V \).

▶️ Answer/Explanation
Solution

Ans: C

Compare \( E = 18,000 – 2,000t \) (linear) and \( V = 18,000 \cdot (0.85^t) \) (exponential) for \( 0 < t < 9 \).

At \( t = 0 \): \( E = 18,000 \), \( V = 18,000 \cdot 0.85^0 = 18,000 \), so \( E = V \).

At \( t = 1 \): \( E = 18,000 – 2,000 = 16,000 \), \( V = 18,000 \cdot 0.85 \approx 15,300 \), so \( E > V \).

At \( t = 9 \): \( E = 18,000 – 2,000 \cdot 9 = 0 \), \( V = 18,000 \cdot 0.85^9 \approx 18,000 \cdot 0.231 \approx 4,158 \), so \( E < V \).

\( E \) starts greater than \( V \) but becomes less than \( V \) by \( t = 9 \).

Choice A: Incorrect, \( E \) is not always less.

Choice B: Incorrect, \( E \) is not always greater.

Choice D: Incorrect, \( E \) starts greater, not less.

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