Question

consider the equation 4y + 3 = 19.
complete the table. tell whether each value of y is a solution or not a solution.
(table with y values and solution? column, y values include 0, 4, 20, 3)

Explanation:

First, let's assume the equation is \( 4v + 3 = 19 \) (since the original is a bit unclear, but this is a common linear equation). To solve for \( v \), we can do:

Step1: Subtract 3 from both sides

\( 4v + 3 - 3 = 19 - 3 \)
\( 4v = 16 \)

Step2: Divide both sides by 4

\( \frac{4v}{4} = \frac{16}{4} \)
\( v = 4 \)

Now, let's check each value (assuming the values are, say, 0, 3, 4, 5; since the table is unclear, but let's use typical values). Wait, maybe the original equation is \( 4v + 3 = 19 \). Let's check each \( v \):

• For \( v = 0 \): \( 4(0) + 3 = 3

eq 19 \) → Not a solution.

• For \( v = 3 \): \( 4(3) + 3 = 15

eq 19 \) → Not a solution.

• For \( v = 4 \): \( 4(4) + 3 = 19 \) → Solution.
• For \( v = 5 \): \( 4(5) + 3 = 23

eq 19 \) → Not a solution.

But since the table is partially visible, let's assume the equation is \( 4v + 3 = 19 \) (common problem). Let's proceed with the values in the table (even if unclear, let's use the solved \( v = 4 \)).

Answer:

Assuming the equation is \( 4v + 3 = 19 \), the solution is \( v = 4 \). For other values:

• If \( v = 0 \): Not a solution.
• If \( v = 3 \): Not a solution.
• If \( v = 4 \): Solution.
• If \( v = 5 \): Not a solution. (Adjust based on actual table values, but the key is solving \( 4v + 3 = 19 \) gives \( v = 4 \))