Question: Solve each equation. Determine whether the equation has one solution, no solution, or infinitely many solutions. (Example 1 and 2)
1. 4(y - 8) + 12 = 2(2y - 9)
2. 9(2x - 5) = 6(x - 4) + 9
3. -4y - 3 = 3(2y - 9) - 8y
4. 6(3 - 5w) = 5(9 - 2w) - 20w
Complete each equation so that it has infinitely many solutions. (Example 3)
5. 2x - 7(x + 10) = x -
6. 12x - x + 8 + 3x = x +
Complete each equation so that it has no solution. (Example 4)
7. -15x + 4x + 2 - x = x +
8. 9(x - 4) - 5x = x -
Test Practice
9. Multiple Choice Which of the following explains why 2/3 k + 3 = 2/3 k - 6 has no solution?
A The coefficients are different, and the constants are different.
B The coefficients are the same, and the constants are the same.
C The coefficients are different, and the constants are the same.
D The coefficients are the same, and the constants are different.
Graph/shape: no graph shapes visible; the page shows worksheet text and handwritten scribble at the lower edge. position_hint: centerUser: help me show your work
1. **Solve** $4(y - 8) + 12 = 2(2y - 9)$
Start by expanding both sides:
$$4y - 32 + 12 = 4y - 18$$
Simplify the left side:
$$4y - 20 = 4y - 18$$
Subtract $4y$ from both sides:
$$\cancel{4y} - 20 = \cancel{4y} - 18$$
$$-20 = -18$$
This is a false statement, so there is **no solution**.
2. **Solve** $9(2x - 5) = 6(x - 4) + 9$
Expand both sides:
$$18x - 45 = 6x - 24 + 9$$
Simplify the right side:
$$18x - 45 = 6x - 15$$
Subtract $6x$ from both sides:
$$18x - 6x - 45 = \cancel{6x} - 15$$
$$12x - 45 = -15$$
Add $45$ to both sides:
$$12x - 45 + 45 = -15 + 45$$
$$12x = 30$$
Divide both sides by $12$:
$$x = \frac{30}{12} = \frac{5}{2}$$
So, there is **one solution**: $x = \frac{5}{2}$.
3. **Solve** $-4y - 3 = 3(2y - 9) - 8y$
Expand the right side:
$$-4y - 3 = 6y - 27 - 8y$$
Simplify the right side:
$$-4y - 3 = -2y - 27$$
Add $2y$ to both sides:
$$-4y + 2y - 3 = -2y + 2y - 27$$
$$-2y - 3 = -27$$
Add $3$ to both sides:
$$-2y - 3 + 3 = -27 + 3$$
$$-2y = -24$$
Divide both sides by $-2$:
$$y = \frac{-24}{-2} = 12$$
So, there is **one solution**: $y = 12$.
4. **Solve** $6(3 - 5w) = 5(9 - 2w) - 20w$
Expand both sides:
$$18 - 30w = 45 - 10w - 20w$$
Simplify the right side:
$$18 - 30w = 45 - 30w$$
Add $30w$ to both sides:
$$18 - 30w + 30w = 45 - 30w + 30w$$
$$18 = 45$$
This is false, so there is **no solution**.
5. **Complete** $2x - 7(x + 10) = x - \,?$ to have infinitely many solutions.
Expand left side:
$$2x - 7x - 70 = x - ?$$
$$-5x - 70 = x - ?$$
For infinitely many solutions, both sides must be identical.
Move $x$ to left:
$$-5x - x - 70 = -?$$
$$-6x - 70 = -?$$
So, right side must be:
$$x - ? = -5x - 70$$
Therefore, $?$ must be $-6x - 70 - x = -70$ (constant part), so the equation is:
$$2x - 7(x + 10) = x - (-70)$$
6. **Complete** $12x - x + 8 + 3x = x + \,?$ to have infinitely many solutions.
Simplify left side:
$$12x - x + 3x + 8 = 14x + 8$$
For infinitely many solutions, right side must equal left side:
$$x + ? = 14x + 8$$
Subtract $x$ from both sides:
$$? = 13x + 8$$
So, complete as:
$$12x - x + 8 + 3x = x + 13x + 8$$
7. **Complete** $-15x + 4x + 2 - x = x + \,?$ to have no solution.
Simplify left side:
$$-15x + 4x - x + 2 = -12x + 2$$
For no solution, the variables must cancel but constants differ.
So, right side must be:
$$x + ? = -12x + c$$
Move $x$ to left:
$$-12x - x + 2 = ?$$
$$-13x + 2 = ?$$
To cancel variables, right side must have $-13x$ term, but since right side is $x + ?$, the $x$ term is positive.
So, no solution if right side is $x + 3$ (constant different from 2).
8. **Complete** $9(x - 4) - 5x = x - \,?$ to have no solution.
Expand left side:
$$9x - 36 - 5x = 4x - 36$$
Simplify left side:
$$4x - 36 = x - ?$$
Move $x$ to left:
$$4x - x - 36 = -?$$
$$3x - 36 = -?$$
For no solution, variables must cancel but constants differ.
So, right side must be $x - 30$ (constant different from 36).
9. **Multiple Choice:** Why does $$\frac{2}{3}k + 3 = \frac{2}{3}k - 6$$ have no solution?
Subtract $\frac{2}{3}k$ from both sides:
$$3 = -6$$
False statement, so coefficients are the same but constants differ.
**Answer: D. The coefficients are the same, and the constants are different.**