Defining New Operations: Practice

Substitute to get an ordinary equation, then solve.

$2x + 4 - 1 = 13 \;\Longrightarrow\; 2x + 3 = 13 \;\Longrightarrow\; 2x = 10 \;\Longrightarrow\; x = 5$

Evaluate both orders.

$3 \oplus 5 = 3 + 5 + 15 = 23, \qquad 5 \oplus 3 = 5 + 3 + 15 = 23$

Both give $23$, so $3 \oplus 5 = 5 \oplus 3$. This rule is symmetric in $a$ and $b$, so it is commutative.

The unknown is the second input, so substitute $a = 3$ and $b = x$.

$2(3) + x - 1 = 12 \;\Longrightarrow\; x + 5 = 12 \;\Longrightarrow\; x = 7$

The unknown is the second input, so substitute $a = 5$ and $b = x$.

$5^2 - x = 20 \;\Longrightarrow\; 25 - x = 20 \;\Longrightarrow\; x = 5$

The average does not depend on the order of the two numbers.

$6 \otimes 10 = \frac{6 + 10}{2} = 8, \qquad 10 \otimes 6 = \frac{10 + 6}{2} = 8$

Both give $8$, so $6 \otimes 10 = 10 \otimes 6$ and the operation is commutative.

Substitute $a = 2$ and $b = y$.

$2 \star y = 2(2) + y - 1 = y + 3$

Substitute $a = x$ and $b = 2$, then solve.

$3x - 2(2) = 11 \;\Longrightarrow\; 3x - 4 = 11 \;\Longrightarrow\; 3x = 15 \;\Longrightarrow\; x = 5$

Evaluate both orders.

$5 \star 8 = 10 + 8 - 1 = 17, \qquad 8 \star 5 = 16 + 5 - 1 = 20$

Since $20 > 17$, the value $8 \star 5$ is larger. The rule doubles the first input, so a larger first input wins.

Substitute $a = x$ and $b = 2$, then combine like terms.

$x \oplus 2 = x + 2 + 2x = 3x + 2$

Substitute $b = 1$ and simplify first.

$x \oplus 1 = x + 1 + x = 2x + 1$

Then solve $2x + 1 = 9$, giving $2x = 8$ and $x = 4$.

Test a pair, such as $2$ and $5$.

$2 \diamond 5 = 4 - 5 = -1, \qquad 5 \diamond 2 = 25 - 2 = 23$

The two orders differ, so $\diamond$ is not commutative. Only the first input is squared, so the slots are not interchangeable.

The unknown is the second input, so substitute $a = 6$ and $b = x$.

$3(6) - 2x = 8 \;\Longrightarrow\; 18 - 2x = 8 \;\Longrightarrow\; -2x = -10 \;\Longrightarrow\; x = 5$