Question

analyzing solution sets to linear equations with the variable on both sides - item 992454
use the drop - down menus to complete each equation so the statement about its solution is true.
no solutions
$5 - 4 + 7x + 1 = \square x + \square$
one solution
$5 - 4 + 7x + 1 = \square x + \square$
infinitely many solutions
$5 - 4 + 7x + 1 = \square x + \square$

Explanation:

First, simplify the left - hand side of the equation \(5 - 4+7x + 1\).

Step 1: Simplify the constant terms

Calculate \(5-4 + 1\). \(5-4=1\), and then \(1 + 1=2\). So the left - hand side simplifies to \(7x+2\).

Case 1: No Solutions

For a linear equation \(ax + b=cx + d\) to have no solutions, we need \(a = c\) and \(b
eq d\). Since the left - hand side is \(7x + 2\), we want the coefficient of \(x\) on the right - hand side to be equal to \(7\) (so \(a = c = 7\)) and the constant term to be different from \(2\). Let's choose the constant term as \(3\) (any number other than 2 will work). So the equation for no solutions is \(5-4 + 7x+1=7x + 3\).

Case 2: One Solution

For a linear equation \(ax + b=cx + d\) to have one solution, we need \(a
eq c\). Since the left - hand side is \(7x + 2\), we can choose the coefficient of \(x\) on the right - hand side to be a number different from \(7\), say \(6\), and the constant term can be any number, say \(2\). So the equation \(5-4 + 7x+1=6x + 2\) will have one solution (we can also choose other values for the coefficient of \(x\) as long as it is not equal to 7).

Case 3: Infinitely Many Solutions

For a linear equation \(ax + b=cx + d\) to have infinitely many solutions, we need \(a = c\) and \(b = d\). Since the left - hand side is \(7x+2\), the equation should be \(5 - 4+7x + 1=7x+2\).

Answer:

• No Solutions: \(5 - 4+7x + 1=\boldsymbol{7}x+\boldsymbol{3}\) (the constant term can be any number except 2, here we choose 3 as an example)
• One Solution: \(5 - 4+7x + 1=\boldsymbol{6}x+\boldsymbol{2}\) (the coefficient of \(x\) can be any number except 7, and the constant term can be any number, here we choose 6 for the coefficient of \(x\) and 2 for the constant term as an example)
• Infinitely Many Solutions: \(5 - 4+7x + 1=\boldsymbol{7}x+\boldsymbol{2}\)