Lesson 15
All, Some, or No Solutions
Let’s think about how many solutions an equation can have.
Problem 1
For each equation, decide if it is always true or never true.
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\(x - 13 = x + 1\)
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\(x+\frac{1}{2} = x - \frac{1}{2}\)
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\(2(x + 3) = 5x + 6 - 3x\)
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\(x - 3 = 2x - 3 -x\)
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\(3(x-5) = 2(x-5) + x\)
Problem 2
Mai says that the equation \(2x + 2 = x +1\) has no solution because the left hand side is double the right hand side. Do you agree with Mai? Explain your reasoning.
Problem 3
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Write the other side of this equation so it's true for all values of \(x\): \(\frac12(6x-10) - x =\)
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Write the other side of this equation so it's true for no values of \(x\): \(\frac12(6x-10) - x = \)
Problem 4
Here is an equation that is true for all values of \(x\): \(5(x+2) = 5x+10\). Elena saw this equation and says she can tell \(20(x+2)+31=4(5x+10)+31\) is also true for any value of \(x\). How can she tell? Explain your reasoning.
Problem 5
Elena and Lin are trying to solve \(\frac12x+3=\frac72x+5\). Describe the change they each make to each side of the equation.
- Elena’s first step is to write \(3=\frac72x-\frac12x+5\).
- Lin’s first step is to write \(x+6=7x+10\).
Problem 6
Solve each equation and check your solution.
\(3x-6=4(2-3x)-8x\)
\(\frac12z+6=\frac32(z+6)\)
\(9-7w=8w+8\)