How do you graph $9 x + 2 y = 5$ $9x+2y=5$ ?

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Answer 1 · 2017-10-27T16:12:03

See a solution process below:

First, solve for two points which solve the equation and plot these points:

First Point: For $x = 1$ $x = 1$

$(9 \cdot 1) + 2 y = 5$ $(9 * 1) + 2y = 5$

$9 + 2 y = 5$ $9 + 2y = 5$

$- 9 + 9 + 2 y = - 9 + 5$ $-color(red)(9) + 9 + 2y = -color(red)(9) + 5$

$0 + 2 y = - 4$ $0 + 2y = -4$

$2 y = - 4$ $2y = -4$

$\frac{2 y}{2} = - \frac{4}{2}$ $(2y)/color(red)(2) = -4/color(red)(2)$

$y = - 2$ $y = -2$ or $(1, - 2)$ $(1, -2)$

Second Point: For $x = - 1$ $x = -1$

$(9 \cdot - 1) + 2 y = 5$ $(9 * -1) + 2y = 5$

$- 9 + 2 y = 5$ $-9 + 2y = 5$

$9 - 9 + 2 y = 9 + 5$ $color(red)(9) - 9 + 2y = color(red)(9) + 5$

$0 + 2 y = 14$ $0 + 2y = 14$

$2 y = 14$ $2y = 14$

$\frac{2 y}{2} = \frac{14}{2}$ $(2y)/color(red)(2) = 14/color(red)(2)$

$y = 7$ $y = 7$ or $(- 1, 7)$ $(-1, 7)$

We can next plot the two points on the coordinate plane:

graph{((x-1)^2+(y+2)^2-0.125)((x+1)^2+(y-7)^2-0.125)=0 [-20, 20, -10, 10]}

Now, we can draw a straight line through the two points to graph the line:

graph{(9x+2y-5)((x-1)^2+(y+2)^2-0.125)((x+1)^2+(y-7)^2-0.125)=0 [-20, 20, -10, 10]}

How do you graph 9x+2y=59x+2y=5?