Question 1 A)
To solve the equation \(2x - 5 = 3(4x + 5)\), follow these steps:
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Distribute on the right side: \[ 2x - 5 = 3 \cdot 4x + 3 \cdot 5 \] \[ 2x - 5 = 12x + 15 \]
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Move all terms involving \(x\) to one side and constant terms to the other side. First, let's subtract \(12x\) from both sides: \[ 2x - 12x - 5 = 15 \] This simplifies to: \[ -10x - 5 = 15 \]
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Now, add \(5\) to both sides: \[ -10x - 5 + 5 = 15 + 5 \] \[ -10x = 20 \]
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Finally, divide by \(-10\): \[ x = \frac{20}{-10} \] \[ x = -2 \]
Final answer: \(x = -2\)
Question 2 A)
To solve the equation \(12 = bY - 3\) for \(Y\), follow these steps:
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Add \(3\) to both sides (to isolate the term containing \(Y\)): \[ 12 + 3 = bY \] \[ 15 = bY \] (Addition Property of Equality)
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Now divide both sides by \(b\) (to solve for \(Y\)): \[ Y = \frac{15}{b} \] (Division Property of Equality)
Final answer: \(Y = \frac{15}{b}\)
Question 3 A)
To derive the equation for the relationship between the number of hours \(h\) Gwen worked and the number of tickets \(t\) she sold, we notice the pattern in the values provided:
- For every hour \(h\), Gwen sold \(23\) tickets.
- Therefore, the total tickets sold \(t\) can be represented as a function of hours worked \(h\):
\[ t = 23h \]
Final answer: \(t = 23h\)
Question 4 A)
To fill in the missing values in the table using the equation \(y - 6 = x \cdot 4\), we can rearrange it to:
\[ y = 4x + 6 \]
Now we can plug in the values for \(x\):
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For \(x = 3\): \[ y = 4(3) + 6 = 12 + 6 = 18 \]
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For \(x = 2\): \[ y = 4(2) + 6 = 8 + 6 = 14 \]
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For \(x = 4\): \[ y = 4(4) + 6 = 16 + 6 = 22 \]
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For \(x = -3\): \[ y = 4(-3) + 6 = -12 + 6 = -6 \]
Final table filled: \[ \begin{array}{|c|c|} \hline x & y \ \hline 3 & 18 \ 2 & 14 \ 4 & 22 \ -3 & -6 \ \hline \end{array} \]
Question 5 A)
To solve the equation \(-8 + 6x - (-12) = 6x - 16 + 2x\):
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Rewrite the equation: \[ -8 + 6x + 12 = 6x - 16 + 2x \]
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Combine like terms: \[ 4 + 6x = 8x - 16 \]
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Move \(6x\) to the right: \[ 4 = 8x - 6x - 16 \] \[ 4 = 2x - 16 \]
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Add \(16\) to both sides: \[ 4 + 16 = 2x \] \[ 20 = 2x \]
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Divide both sides by \(2\): \[ x = 10 \]
Final answer: \(x = 10\)
Question 6 A)
Identify the steps followed to solve the equation \(5 - 3(x + 3) = 11 - 8x\):
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Distributive Property: \[ 5 - 3x - 9 = 11 - 8x \]
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Combine Like Terms: \[ -3x - 4 = 11 - 8x \]
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Addition Property of Equality (adding \(8x\)): \[ 5x - 4 = 11 \]
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Addition Property of Equality (adding \(4\)): \[ 5x = 15 \]
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Division Property of Equality: \[ x = 3 \]
Question 7 A)
Identify the property used to arrive at Line 1:
The equation is \(x + 2(x + 1) = 17\).
In Line 1, we distribute \(2\) through \( (x + 1) \): \[ x + 2x + 2 = 17 \] Property used: Distributive Property
Question 8 A)
Solve the equation: \[ \frac{y}{7} + 8 = 22 \]
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Subtract \(8\) from both sides: \[ \frac{y}{7} = 14 \]
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Multiply both sides by \(7\): \[ y = 98 \]
Final answer: \(y = 98\)
Question 9 A)
Solve \(18x - 3 = 2\):
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Add \(3\): \[ 18x = 5 \]
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Divide by \(18\): \[ x = \frac{5}{18} \]
Which equation has the same solution?
Final Answer: \(x - 3 = 18\)
Question 10 A)
The setup is: \[ x + (x + 2) + (x + 4) = 120 \]
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Combine: \[ 3x + 6 = 120 \]
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Subtract \(6\): \[ 3x = 114 \]
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Divide: \[ x = 38 \]
The integers are \(38\), \(40\), and \(42\), with the largest being \(42\).
Final Answer: Largest integer is \(42\).
Question 11 A)
To solve for \(b\) in the area formula: \[ A = \frac{1}{2}hb \]
Multiply both sides by \(2\): \[ 2A = hb \]
Now divide by \(h\): \[ b = \frac{2A}{h} \]
Final answer: \(b = \frac{2A}{h}\)
Question 12 A)
Solve: \[ 7(5x + 2) = -4(6 - 5x) \]
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Distribute: \[ 35x + 14 = -24 + 20x \]
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Move \(20x\) to the left: \[ 15x + 14 = -24 \]
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Subtract \(14\): \[ 15x = -38 \]
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Divide by \(15\): \[ x = -\frac{38}{15} \]
Final answer: \(x = -\frac{38}{15}\)
Question 13 A)
Set up the equation: \[ 3(r + 10) + 3r = 282 \]
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Distribute: \[ 3r + 30 + 3r = 282 \]
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Combine: \[ 6r + 30 = 282 \]
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Subtract \(30\): \[ 6r = 252 \]
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Divide by \(6\): \[ r = 42 \]
Final answer: \(r = 42\) mph
Question 14 A)
To show the triangle area formula in terms of \(b\): \[ A = \frac{1}{2}hb \]
Multiply by \(2\): \[ 2A = hb \]
Now divide by \(h\): \[ b = \frac{2A}{h} \]
Final answer: \(b = \frac{2A}{h}\)
Question 15 A)
Solve the system of inequalities:
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From \(2x + 3 \geq 7\):
- Subtract \(3\): \[ 2x \geq 4 \]
- Divide by \(2\): \[ x \geq 2 \]
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From \(2x \geq 8\): \[ x \geq 4 \]
Final answer: The solution set is \( {x | x \geq 4} \).
Question 16 A)
Solve the equations:
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From \(3x - 2 = 7\): \[ 3x = 9 \] \[ x = 3 \]
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From \(4x + 5 \leq 21\): \[ 4x \leq 16 \] \[ x \leq 4 \]
Final answer: The solution set is \( {x | x \leq 4} \).