✒️ Equation
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A. To find the x - intercept, let y = 0.
![\tt \: \: \large \tt \: \begin{array}c \hline \tt \: 4x \: - \: 3y \: = \: \tt\underline{\green{ 12 }} \\ \tt 4x \: - \: 3(0) \: = \: \tt\underline{\blue{ 12 }} \\ \tt4x \: - \: 0 \: = \: \tt\underline{\pink{ 12 }} \\ \tt \tt\underline{\red{ \: x \: = \: 3 }} \\ \\ \tt \: x ↬\: the \: intercept\\ \tt\\ \hline\end{array} \: \\]()
To find the y - intercept, let x = 0
![\large \tt \: \begin{array} \hline \tt 4x \: - \: 3y \: = \: \tt\underline{\green{ 12 }} \\ \tt 4(0) \: - \: 3y \: = \: \tt\underline{\blue{ 12 }} \\ \tt \: 0 \: - \: 3y \: = \: \tt\underline{\pink{ 12 }} \\ \tt \tt\underline{\red{ y \: = \: - 4 }} \\ \\ \tt \: x ↬\: the \: intercept \\ \ \\ \hline\end{array} \:]()
B. Replace x with 0 and solve for y. Then replace y with 0 and solve for x.
![\large \tt \: \begin{array}c \hline \tt y \: - \: intercept: \: y \: = \: (1)/(3)(0) \\ \quad \quad \quad \quad \quad \quad \quad \quad \quad \tt = \tt\underline{\red{ 0 }} \\ \\ \tt \: x \: and \: y ↬ intercepts: (0,0)\\ \\ \tt \: x \: ↬ \: intercepts: \: 0 = \: (1)/(3) x \\ \quad\quad \quad \quad\quad\quad\quad\quad\quad \: \tt \tt\underline{\red{ \: 0 \: = \: x }} \ \\ \\ \hline\end{array} \\]()
The equation in b is a linear equation in the form y = mx, where m is any real number. the graph of any equation in the form y = mx is always a line that passes through the origin, which means that both the x - and - y - intercepts are (0,0).
![\large \tt \: \begin{array} \hline \tt intercept \: for \: m \: of \: y \: = \: \tt\underline{\red{ mx }} \\ \hline\end{array}]()
C. Replace y with 0, then solve for the x.
![\large \tt \: \begin{array}c \hline \tt x ↬ \: intercept : \: 0 \: = \: 2x \: \ + 5 \\ \\ \tt - 5 \: = \: 2x\\ \\ \tt \: ( - 5)/(2) \: = \: x\\ \\ \tt \: x \: ↬ \: intercept \ \\ \hline\end{array}]()
Now, replace x with 0, and then solve for y.
![\large \tt \: \begin{array} \hline \tt y ↬ \: intercept: \: y \: = \: 2(0) \: + \: 5 \\ \\ \tt \: y \: = \: 0 \: + \: 5 \\ \\ \tt \: y \: = \: 5 \\ \\ \tt y ↬ \: intercept \ \\ \\ \hline\end{array}]()
The equation in 11c is written in the form y = mx + b. note than in the form, when x is replaced by 0 to find the y↬intercept. you are left with the constant b.
![\large \tt \: \begin{array} \hline \tt \: the \: y ↬ intercept \: of \: y \: = \: mx \: + \: b \\ \hline\end{array}]()
If the equation is in the form y = mx + b, where m and b are real numbers, then the y ↬intercept is b.
D. Remember that the graph of y = 6 is horizontally line parallel to the x axis that passes through the y axis at (0,6). since the line is parallel to x axis, it will not incest tha x-axis. Hence, there is no x ↬ intercept.
![\pmb{(ノ‥)ノ}\qquad\qquad\qquad\qquad\boxed{\tt{[05-12-2022]}} \\ \qquad\qquad\qquad\qquad\qquad\qquad\boxed{\tt{[03:11 \: am]}}]()