(a) The linear equation is y=-1+2x.

In a linear equation \(\displaystyle{y}={b}_{{0}}+{b}_{{1}}{x}\) the constant \(\displaystyle{b}_{{1}}\) be the slope and \(\displaystyle{b}_{{0}}\) be the y-intercept form and x is the independent variable and y is the independent variable.

Comparing the given equation with the general form of linear equation the slope of the equation is 4 and the y-intercept is 3. Thus, the slope of the linear equation is \(\displaystyle{b}_{{1}}={2}\) and the y-intercept is \(\displaystyle{b}_{{0}}=-{1}\).

(b) In a linear equation \(\displaystyle{y}={b}_{{0}}+{b}_{{1}}{x}\) the constant \(\displaystyle{b}_{{1}}\) be the slope and \(\displaystyle{b}_{{0}}\) be the y-intercept form and x is the independent variable and y is the independent variable.

It is known that, the slope of the linear equation \(\displaystyle{y}={b}_{{0}}+{b}_{{1}}{x}\) is upward if \(\displaystyle{b}_{{1}}{>}{0}\), the slope of the linear equation \(\displaystyle{y}={b}_{{0}}+{b}_{{1}}{x}{y}={b}_{{0}}+{b}_{{1}}{x}\) is downward if \(\displaystyle{b}_{{1}}{<}{0}\)</span> and the slope of the linear equation \(\displaystyle{y}={b}_{{0}}+{b}_{{1}}{x}\) is horizontal if \(\displaystyle{b}_{{1}}={0}.\)

Thus, in the given equation \(\displaystyle{y}=-{1}+{2}{x},{b}_{{1}}={2}{<}{0}.\)</span>

Thus, the slope is upward.

(c) The linear equation is y=-1+2x.

Graph the line:

The two points \(\displaystyle{\left({x}_{{1}}, {y}_{{1}}\right)}{\quad\text{and}\quad}{\left({x}_{{2}}, {y}_{{2}}\right)}\) on the given line are obtained:

If x = 0,

\(\displaystyle{y}=-{1}+{\left({2}\times{0}\right)}\)

y=-1

Thus, one point on the line is \(\displaystyle{\left({x}_{{1}}, {y}_{{1}}\right)}={\left({0},-{1}\right)}.\)

If x = 1

\(\displaystyle{y}=-{1}+{\left({2}\times{1}\right)}\)

y=1

Thus, the second point on the line is (x_2, y_2)=(1,1).

In a linear equation \(\displaystyle{y}={b}_{{0}}+{b}_{{1}}{x}\) the constant \(\displaystyle{b}_{{1}}\) be the slope and \(\displaystyle{b}_{{0}}\) be the y-intercept form and x is the independent variable and y is the independent variable.

Comparing the given equation with the general form of linear equation the slope of the equation is 4 and the y-intercept is 3. Thus, the slope of the linear equation is \(\displaystyle{b}_{{1}}={2}\) and the y-intercept is \(\displaystyle{b}_{{0}}=-{1}\).

(b) In a linear equation \(\displaystyle{y}={b}_{{0}}+{b}_{{1}}{x}\) the constant \(\displaystyle{b}_{{1}}\) be the slope and \(\displaystyle{b}_{{0}}\) be the y-intercept form and x is the independent variable and y is the independent variable.

It is known that, the slope of the linear equation \(\displaystyle{y}={b}_{{0}}+{b}_{{1}}{x}\) is upward if \(\displaystyle{b}_{{1}}{>}{0}\), the slope of the linear equation \(\displaystyle{y}={b}_{{0}}+{b}_{{1}}{x}{y}={b}_{{0}}+{b}_{{1}}{x}\) is downward if \(\displaystyle{b}_{{1}}{<}{0}\)</span> and the slope of the linear equation \(\displaystyle{y}={b}_{{0}}+{b}_{{1}}{x}\) is horizontal if \(\displaystyle{b}_{{1}}={0}.\)

Thus, in the given equation \(\displaystyle{y}=-{1}+{2}{x},{b}_{{1}}={2}{<}{0}.\)</span>

Thus, the slope is upward.

(c) The linear equation is y=-1+2x.

Graph the line:

The two points \(\displaystyle{\left({x}_{{1}}, {y}_{{1}}\right)}{\quad\text{and}\quad}{\left({x}_{{2}}, {y}_{{2}}\right)}\) on the given line are obtained:

If x = 0,

\(\displaystyle{y}=-{1}+{\left({2}\times{0}\right)}\)

y=-1

Thus, one point on the line is \(\displaystyle{\left({x}_{{1}}, {y}_{{1}}\right)}={\left({0},-{1}\right)}.\)

If x = 1

\(\displaystyle{y}=-{1}+{\left({2}\times{1}\right)}\)

y=1

Thus, the second point on the line is (x_2, y_2)=(1,1).