Let's go through each equation step-by-step to determine if it is in the slope-intercept form of a linear equation, [tex]\( y = mx + b \)[/tex], where [tex]\( m \)[/tex] is the slope and [tex]\( b \)[/tex] is the y-intercept.
1. Equation 1: [tex]\( y = 6x - 5 \)[/tex]
- This is in the form [tex]\( y = mx + b \)[/tex] where [tex]\( m = 6 \)[/tex] and [tex]\( b = -5 \)[/tex].
- Therefore, this equation is in slope-intercept form.
2. Equation 2: [tex]\( y = -\frac{7}{4}x + 2 \)[/tex]
- This is in the form [tex]\( y = mx + b \)[/tex] where [tex]\( m = -\frac{7}{4} \)[/tex] and [tex]\( b = 2 \)[/tex].
- Therefore, this equation is in slope-intercept form.
3. Equation 3: [tex]\( y = x^2 + 3 \)[/tex]
- This equation has an [tex]\( x^2 \)[/tex] term, which makes it a quadratic equation, not a linear equation.
- Therefore, this equation is not in slope-intercept form as it does not meet the criteria for [tex]\( y = mx + b \)[/tex].
4. Equation 4: [tex]\( y + 3 = x \)[/tex]
- Rearranging this equation, we get [tex]\( y = x - 3 \)[/tex], which is in the form [tex]\( y = mx + b \)[/tex] where [tex]\( m = 1 \)[/tex] and [tex]\( b = -3 \)[/tex].
- Therefore, this equation is in slope-intercept form.
Based on our analysis, the table should look like this:
\begin{tabular}{|c|c|c|c|}
\hline
& Slope-Intercept Form & Not Slope-Intercept Form \\
\hline
[tex]$y=6 x-5$[/tex] & O & \\
\hline
[tex]$y=-\frac{7}{4} x+2$[/tex] & O & \\
\hline
[tex]$y=x^2+3$[/tex] & & O \\
\hline
[tex]$y+3=x$[/tex] & O & \\
\hline
\end{tabular}
The correct determination for each equation is:
[tex]\[ [1, 1, 0, 1] \][/tex]
