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Use the Substitution Method, and then use slope and y-intercept to graph the equations.
Use the Substitution Method, and then use slope and y-intercept to graph the equations.
(1,2)
(-3,2)
Let's start with solving the system algebraically. We will use the Substitution Method to solve this system of equations. It is usually the best choice when one of the variables is already isolated or has a coefficient of 1 or -1. In the second equation, x is already solved for, so we can substitute it in the first equation to find y.
Having found y, we can substitute this into the second equation to find x.
We can confirm our solution by graphing the equations on the same set of axes. To do this we should rewrite them into the slope-intercept form.
| Given Equation | Slope-Intercept Form | Slope m | y-intercept b |
|---|---|---|---|
| x+y=3 | y=-x+ 3 | -1 | (0, 3) |
| x=3y-5 | y=1/3x+ 5/3 | 1/3 | (0, 5/3) |
To graph each line, we will plot the y-intercept first, then we will use the slope to find another point. Connecting the points will draw the line for each equation.
The lines intersect at (1,2), so we confirmed that our solution is correct.
Again we will start with solving the system algebraically. Since in the second equation y is already solved for, we will use the Substitution Method to solve this system of equations.
Having found x, we can substitute this into the second equation to find y.
(II): x= -3
(II): - a(- b)=a* b
(II):Subtract term
We can confirm our solution by graphing the equations on the same set of axes. To do this we should rewrite them into the slope-intercept form.
| Given Equation | Slope-Intercept Form | Slope m | y-intercept b |
|---|---|---|---|
| x-y=-5 | y=x+ 5 | 1 | (0, 5) |
| y=-2x-4 | y=-2x+( -4) | -2 | (0, -4) |
To graph each line, we will plot the y-intercept first, then we will use the slope to find another point. Connecting the points will draw the line for each equation.
The lines intersect at (-3,2), so we confirmed that our solution is correct.