Question

determine whether each equation has no solution, one solution, or infinitely many solutions. drag each equation to the correct location on the table. no solution one solution infinitely many solutions -1.7v + 2.8 = 1.4v - 3.1v + 2.8 4a - 3 + 2a = 7a - 2 (1/5)f - (2/3) = - (1/5)f + (2/3) 2y - 3 = 5 + 2(y - 1) -3(n + 4) + n = -2(n + 6)

Explanation:

To determine the number of solutions for each equation, we simplify and analyze each one:

Equation 1: \(-1.7v + 2.8 = 1.4v - 3.1v + 2.8\)

• Simplify the right - hand side: \(1.4v-3.1v=-1.7v\), so the equation becomes \(-1.7v + 2.8=-1.7v + 2.8\).
• Subtract \(-1.7v + 2.8\) from both sides, we get \(0 = 0\). This is always true, so the equation has infinitely many solutions.

Equation 2: \(4a-3 + 2a=7a - 2\)

• Combine like terms on the left - hand side: \(4a + 2a=6a\), so the equation is \(6a-3 = 7a-2\).
• Subtract \(6a\) from both sides: \(-3=a - 2\).
• Add 2 to both sides: \(a=-1\). There is one solution.

Equation 3: \(\frac{1}{5}f-\frac{2}{3}=-\frac{1}{5}f+\frac{2}{3}\)

• Add \(\frac{1}{5}f\) to both sides: \(\frac{1}{5}f+\frac{1}{5}f-\frac{2}{3}=\frac{2}{3}\), \(\frac{2}{5}f-\frac{2}{3}=\frac{2}{3}\).
• Add \(\frac{2}{3}\) to both sides: \(\frac{2}{5}f=\frac{2}{3}+\frac{2}{3}=\frac{4}{3}\).
• Multiply both sides by \(\frac{5}{2}\): \(f=\frac{4}{3}\times\frac{5}{2}=\frac{10}{3}\). There is one solution.

Equation 4: \(2y-3 = 5+2(y - 1)\)

• Expand the right - hand side: \(2y-3 = 5 + 2y-2\).
• Simplify the right - hand side: \(2y-3=2y + 3\).
• Subtract \(2y\) from both sides: \(-3 = 3\), which is false. So the equation has no solution.

Equation 5: \(-3(n + 4)+n=-2(n + 6)\)

• Expand both sides: \(-3n-12 + n=-2n-12\).
• Combine like terms on the left - hand side: \(-2n-12=-2n-12\).
• Subtract \(-2n - 12\) from both sides, we get \(0 = 0\). This is always true, so the equation has infinitely many solutions.

Final Table:

No Solution	One Solution	Infinitely Many Solutions

Answer:

To determine the number of solutions for each equation, we simplify and analyze each one:

Equation 1: \(-1.7v + 2.8 = 1.4v - 3.1v + 2.8\)

• Simplify the right - hand side: \(1.4v-3.1v=-1.7v\), so the equation becomes \(-1.7v + 2.8=-1.7v + 2.8\).
• Subtract \(-1.7v + 2.8\) from both sides, we get \(0 = 0\). This is always true, so the equation has infinitely many solutions.

Equation 2: \(4a-3 + 2a=7a - 2\)

• Combine like terms on the left - hand side: \(4a + 2a=6a\), so the equation is \(6a-3 = 7a-2\).
• Subtract \(6a\) from both sides: \(-3=a - 2\).
• Add 2 to both sides: \(a=-1\). There is one solution.

Equation 3: \(\frac{1}{5}f-\frac{2}{3}=-\frac{1}{5}f+\frac{2}{3}\)

• Add \(\frac{1}{5}f\) to both sides: \(\frac{1}{5}f+\frac{1}{5}f-\frac{2}{3}=\frac{2}{3}\), \(\frac{2}{5}f-\frac{2}{3}=\frac{2}{3}\).
• Add \(\frac{2}{3}\) to both sides: \(\frac{2}{5}f=\frac{2}{3}+\frac{2}{3}=\frac{4}{3}\).
• Multiply both sides by \(\frac{5}{2}\): \(f=\frac{4}{3}\times\frac{5}{2}=\frac{10}{3}\). There is one solution.

Equation 4: \(2y-3 = 5+2(y - 1)\)

• Expand the right - hand side: \(2y-3 = 5 + 2y-2\).
• Simplify the right - hand side: \(2y-3=2y + 3\).
• Subtract \(2y\) from both sides: \(-3 = 3\), which is false. So the equation has no solution.

Equation 5: \(-3(n + 4)+n=-2(n + 6)\)

• Expand both sides: \(-3n-12 + n=-2n-12\).
• Combine like terms on the left - hand side: \(-2n-12=-2n-12\).
• Subtract \(-2n - 12\) from both sides, we get \(0 = 0\). This is always true, so the equation has infinitely many solutions.

Final Table:

No Solution	One Solution	Infinitely Many Solutions