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Algebra / Systems of two linear equations in two variables Difficulty: Hard

$6 + 7 r = p w$

$7 r - 5 w = 5 w + 11$

In the given system of equations, $p$ is a constant. If the system has no solution, what is the value of $p$ ?

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Explanation

The correct answer is $10$ . Solving by substitution, the given system of equations, where $p$ is a constant, can be written so that the left-hand side of each equation is equal to $7 r$ . Subtracting $6$ from each side of the first equation in the given system, $6 + 7 r = p w$ , yields $7 r = p w - 6$ . Adding $5 w$ to each side of the second equation in the given system, $7 r - 5 w = 5 w + 11$ , yields $7 r = 10 w + 11$ . Since the left-hand side of each equation is equal to $7 r$ , setting the the right-hand side of the equations equal to each other yields $p w - 6 = 10 w + 11$ . A linear equation in one variable, $w$ , has no solution if and only if the equation is false; that is, when there's no value of $w$ that produces a true statement. For the equation $p w - 6 = 10 w + 11$ , there's no value of $w$ that produces a true statement when $p w equals 10 w$ . Therefore, for the equation $p w - 6 = 10 w + 11$ , there's no value of $w$ that produces a true statement when the value of $p$ is $10$ . It follows that in the given system of equations, the system has no solution when the value of $p$ is $10$ .