- Simplify both sides first. - Decide what it means if both sides become the same expression.

1. Distribute the left side: \(4x + 8 = 4x + 8\). 2. Both sides are the same expression, so the equation is true for every value of \(x\). 3. Therefore, the equation has infinitely many solutions.

Infinitely many solutions

- Compare the variable terms on both sides. - Remove matching variable terms and interpret the statement that remains.

1. Subtract \(3x\) from both sides: \(5 = -2\). 2. This statement is false, so no value of \(x\) can make the original equation true. 3. Therefore, the equation has no solution.

No solution

- What value solves the first equation? - What happens to the variable terms in the second equation?

1. In a), divide by \(2\), so only \(x=0\) works. 2. In b), subtract \(3x\) from both sides and get \(4=5\), which is impossible.

a) Only \(x=0\). b) No solution.

- Simplify each equation as far as possible. - A contradiction means no solution. - An identity means infinitely many solutions. - Do not confuse \(x = 0\) with having no solution.

1. The first equation gives \(3=-2\) after subtracting \(5x\). This contradiction means there is no solution. 2. The second equation simplifies to \(2x+8=2x+8\), an identity true for every real value of \(x\), so there are infinitely many solutions. 3. The third equation gives \(x=0\) after dividing by \(3\), so it has exactly one solution.

(1) No solution (2) Infinitely many solutions (3) Exactly one solution, \(x = 0\)

- Look for properties such as combining like terms, the distributive property, or the identity property. - A single counterexample proves that an equation is not an identity. - Try rewriting one side until it matches the other, or solve the equation to see whether only certain values work.

1. Combining like terms shows \(n+n=2n\) for every \(n\), so the first equation is an identity. 2. The equation \(3x=15\) is true only when \(x=5\), so the second equation is not an identity. 3. The distributive property gives \(4(a+b)=4a+4b\) for all allowable values, so the third equation is an identity. 4. Solving \(y-5=5-y\) gives \(2y=10\), so \(y=5\). The fourth equation is not an identity. 5. The multiplicative identity property gives \(z\cdot1=z\) for every \(z\), so the fifth equation is an identity.

Statements 1), 3), and 5) are identities.

- Simplify each equation until any matching variable terms have canceled. - Does the remaining statement have to be true, false, or true only for a particular value? - Can an equation have exactly one solution even if that solution is zero?

1. Equation 1 gives \(4x-4=4x-1\), then \(-4=-1\), which is false. It has no solution. 2. Equation 2 gives \(2x+8=2x+8\), so all values of \(x\) work. 3. Equation 3 gives \(5=7\) after adding \(x\) to both sides, which is false. It has no solution. 4. Equation 4 gives \(x=0\), a valid solution. 5. Only equations 1 and 3 have no solution.

1) No solution: \(-4=-1\) is false. 3) No solution: \(5=7\) is false.

- Solve every equation separately. - Write or compare the resulting solution sets. - Equivalent equations must have exactly the same solutions.

1. For the first pair, \(4x-7=13\) gives \(x=5\), and \(2x+5=15\) also gives \(x=5\). Both equations have solution set \(S=\{5\}\), so they are equivalent. 2. For the second pair, \(3(x-2)=12\) gives \(x=6\), while \(x+4=9\) gives \(x=5\). The solution sets differ, so the equations are not equivalent.

a) Equivalent; both have solution set \(S = \{5\}\). b) Not equivalent; the first has \(S = \{6\}\) and the second has \(S = \{5\}\).

- Rewrite the parentheses in each equation. - What remains when the variable terms cancel? - Decide whether the remaining statement is always true or impossible.

1. Distribute the left side in both equations: \(4x+4\). 2. In a), the sides are identical, so every \(x\) works. 3. In b), subtracting \(4x\) gives \(4=5\), so no \(x\) works.

a) Infinitely many solutions. b) No solution.

- What does the original equation become after distribution? - Try two different values of \(x\) in the original equation. - Does Morgan's last step preserve all values that worked before it?

1. Distributing gives \(6x+12=6x+12\). Both sides are identical for every value of \(x\). 2. Dividing by \(0\) is not allowed. The equation has infinitely many solutions, not only \(x=0\).

Division by \(0\) is invalid. Every value of \(x\) is a solution.

- A true statement such as \(0 = 0\) means every value in the domain works. - A false statement such as \(0 = 5\) means no value works. - Distribute and combine like terms before deciding the number of solutions.

1. The first equation simplifies to \(3x+3=3x+3\), an identity true for every rational number, so \(S=\mathbb{Q}\). 2. The second equation gives \(5=-1\) after subtracting \(2x\), a contradiction, so \(S=\varnothing\). 3. The third equation becomes \(4x-10=2x+2\), so \(2x=12\), \(x=6\), and \(S=\{6\}\). 4. The fourth equation gives \(\frac{x}{4}=3\), so \(x=12\) and \(S=\{12\}\).

a) Infinitely many solutions: \(S = \mathbb{Q}\) b) No solution: \(S = \varnothing\) c) One solution: \(S = \{6\}\) d) One solution: \(S = \{12\}\)

- Distribute and combine like terms first. - Notice what happens if the variable terms cancel. - A false numerical statement means there is no solution. - Use the empty-set symbol for the solution set.

1. Distribute: \(6n - 12 - 2n - 2 = 4n + 12\). 2. Combine like terms: \(4n - 14 = 4n + 12\). 3. Subtract \(4n\) from both sides: \(-14 = 12\). 4. This contradiction is never true, so the equation has no solution and \(S = \varnothing\).

The equation has no solution, so \(S = \varnothing\).

- Think about what must happen after simplifying when an equation is true for every input. - Contrast that with a simplification that produces an impossible statement. - For a unique prescribed solution, make the two sides agree at that input without becoming identical expressions.

1. For infinitely many solutions, both sides must be identical. One valid right side is \(4x+10\). 2. For no solution, keep the same variable term but change the constant. For example, \(4x+1\) gives \(10=1\) after subtracting \(4x\). 3. To make \(x=5\) the unique solution, evaluate the left side at \(5\): \(4\cdot5+10=30\). Using the right side \(30\) gives \(4x+10=30\), whose only solution is \(5\).

a) One answer is \(4x+10\). b) One answer is \(4x+1\). c) One answer is \(30\).

- Zero is a valid number. - Solve both equations and compare the resulting statements. - Distinguish the set \(\{0\}\) from the empty set \(\varnothing\).

1. For \(8x = 0\), divide by \(8\): \(x = 0\). Zero is a number and is a valid solution, so \(S = \{0\}\). 2. For \(x + 1 = x\), subtract \(x\) from both sides: \(1 = 0\). This contradiction is never true, so \(S = \varnothing\). 3. Jonah’s claim is incorrect. A solution of zero means there is exactly one solution; an empty solution set means there is no solution.

Jonah is incorrect. For \(8x = 0\), \(S = \{0\}\). For \(x + 1 = x\), \(S = \varnothing\).

- Compare the slopes of the two lines in each equation. - Simplify any expression with parentheses before comparing. - Recall when two lines are parallel and when they coincide.

1. For the first equation, the lines \(y=2x+5\) and \(y=2x-3\) have the same slope but different \(y\)-intercepts. They are distinct parallel lines, so there is no solution. 2. For the second equation, simplify the right side: \(\frac{1}{2}(6x-2)=3x-1\). Both sides represent the same line, so every real value of \(x\) is a solution. 3. For the third equation, the lines \(y=4x+2\) and \(y=-x+7\) have different slopes. They intersect once, so the equation has exactly one solution.

1. No solution; the lines are parallel and distinct. 2. Infinitely many solutions; both expressions represent the same line. 3. Exactly one solution; the lines have different slopes.

- Simplify both sides before deciding how many values work. - What happens if every variable term cancels? - When a variable remains, solve to see whether a single value works.

1. For part a, subtract \(2x\) and \(7\): \(3x=-15\), so \(x=-5\). There is one solution. 2. For part b, distribute: \(6x+12=6x-1\). Subtract \(6x\): \(12=-1\), a contradiction. There is no solution. 3. For part c, distribute: \(3x+6=2x+10\). Thus \(x=4\), so there is one solution.

a) One solution: \(x=-5\). b) No solution. c) One solution: \(x=4\).

- Simplify both sides before comparing them. - Handle minus signs before parentheses carefully. - An identity has infinitely many solutions. - A contradiction has no solution.

1. For equation A, simplify both sides: \(4x-2x-6=2x-6\), so \(2x-6=2x-6\). This identity has infinitely many solutions. 2. For equation B, distribute on the right: \(\frac{2}{3}x+5=\frac{2}{3}x+4\). Subtracting \(\frac{2}{3}x\) gives \(5=4\), so there is no solution. 3. For equation C, simplify the right side: \(3x+7=4x-3\). Solving gives \(x=10\), so there is exactly one solution.

A: Infinitely many solutions B: No solution C: Exactly one solution

- Substitute each value of \(k\) separately. - Simplify the right side first. - Compare the variable coefficients and constants after simplification.

1. Simplify the right side to \(5x+12\). 2. With \(k=5\), the equation is \(5x+12=5x+12\), an identity, so there are infinitely many solutions. 3. With \(k=4\), the equation is \(4x+12=5x+12\). Subtracting \(4x\) and \(12\) gives \(x=0\), so there is exactly one solution. 4. With \(k=5\) and the left constant changed to \(10\), the equation is \(5x+10=5x+12\). Subtracting \(5x\) gives \(10=12\), so there is no solution.

a) Infinitely many solutions b) Exactly one solution, \(x = 0\) c) No solution

- Simplify both sides completely. - Notice what happens when the variable terms cancel. - A false numerical statement means there is no solution.

1. Remove the parentheses on the left: \(6x - 2x - 4 = 4(x + 1)\). 2. Simplify and distribute: \(4x - 4 = 4x + 4\). 3. Subtract \(4x\): \(-4 = 4\). 4. This contradiction is never true, so \(S = \varnothing\).

\(S = \varnothing\)

- Simplify the left side and compare it with the right side. - An identity is true for every value in the stated domain. - A solution set cannot include values outside its domain.

1. Distribute on the left: \(0.5(8x-4)=4x-2\), so the equation becomes \(4x-2=4x-2\), an identity. 2. Over the rational-number domain, every allowable value works, so \(S=\mathbb{Q}\). 3. Over the finite domain \(\{0,1,2\}\), the identity is true for every listed value, so \(S=\{0,1,2\}\).

a) \(S = \mathbb{Q}\) b) \(S = \{0, 1, 2\}\)

- In each equation, compare what remains after the same variable quantity is removed from both sides. - Decide whether the resulting statement is false or always true. - Connect a contradiction with no solutions and an identity with all real-number solutions.

1. For Example A, subtract \(3x\) from both sides: \(4 = 9\). This contradiction is false, so the equation has no solution. 2. For Example B, subtract \(3x\) from both sides: \(4 = 4\). This identity is always true, so every real number is a solution. 3. Njeri's claim is incorrect. Equal variable terms can lead to no solution when the constants differ, or infinitely many solutions when the constants are equal.

The claim is incorrect. Example A has no solution. Example B is true for every real number, so every real number is a solution.

- Equivalent equations have the same solution set. - Look for a reversible operation applied to both sides. - Apply an operation to every term on each side. - Check whether the proposed second equation is exactly the result of the transformation.

1. Adding \(4\) to both sides of the first equation in the first pair gives \(x=16\), so the equations are equivalent. 2. Dividing both sides of \(6x=24\) by \(3\) gives \(2x=8\), so the second pair is equivalent. 3. Adding \(5\) to both sides of \(x+5=10\) gives \(x+10=15\), not \(x+10=20\), so the third pair is not equivalent. 4. Multiplying every term of \(\frac{x}{2}+3=7\) by \(2\) gives \(x+6=14\), so the fourth pair is equivalent.

1) Equivalent; add \(4\) to both sides. 2) Equivalent; divide both sides by \(3\). 3) Not equivalent. 4) Equivalent; multiply every term by \(2\).

- Ask when the variable terms cancel after moving them to one side. - A true statement such as \(0 = 0\) gives infinitely many solutions. - A false statement such as \(0 = 1\) gives no solution. - For part c, substitute \(x = 5\) and choose constants that make the equation true.

1. For infinitely many solutions, both sides must be identical. One choice is \(a=4\) and \(b=12\). 2. For no solution, the coefficients of \(x\) must match while the constants differ. One choice is \(a=4\) and \(b=10\). 3. To make \(x=5\) the unique solution, substitute \(5\): \(4\cdot5+12=5a+b\), so \(32=5a+b\). Also require \(a\ne4\). 4. Choosing \(a=2\) gives \(b=22\), so \(a=2\), \(b=22\) is one valid example.

a) One example is \(a = 4\), \(b = 12\). b) One example is \(a = 4\), \(b = 10\). c) One example is \(a = 2\), \(b = 22\).

- Distribute on the right side first. - Infinitely many solutions occur when both sides simplify to the same expression. - One solution occurs when the variable coefficients differ. - No solution occurs when equal variable terms leave unequal constants.

1. Distribute on the right: \(4(2x+\square)=8x+4\square\). To match \(8x+20\), require \(4\square=20\), so the box value is \(5\). 2. For exactly one solution, choose a left-side coefficient different from \(8\). For example, \(7x+20=4(2x+5)\) becomes \(7x+20=8x+20\), so \(x=0\). 3. For no solution, keep equal variable coefficients but make the constants different. For example, \(8x+20=4(2x+6)\) becomes \(8x+20=8x+24\), a contradiction.

a) \(5\) b) One example is \(7x + 20 = 4(2x + 5)\), which has solution \(x = 0\). c) One example is \(8x + 20 = 4(2x + 6)\), which has no solution.