Solve by either graphing and then get an exact solution by using either substitution or elimination:
x-3y=6
2x+y=2
both are straight lines. For x = 0 find y for both equations and mark those points on a graph
Then for x = something else, like 10, find y for both equations
draw lines through the points for each equation
Find where the lines cross.
x = (3y+6)
substitute that in the second one
2(3y+6) + y = 2
plz help me
To solve this system of equations, we will start by graphing the equations to find a rough estimate of the solution. Then, we will use either substitution or elimination to find the exact solution.
1. Graphing:
To graph the equations, we can rewrite them in slope-intercept form (y = mx + b), where m is the slope and b is the y-intercept.
Equation 1: x - 3y = 6
Rewrite it as: -3y = -x + 6, then divide by -3: y = (1/3)x - 2
Equation 2: 2x + y = 2
Rewrite it as: y = -2x + 2
Now, plot the graphs of these two lines on the same coordinate plane:
The line represented by y = (1/3)x - 2 has a y-intercept of -2 and a slope of (1/3). Draw a line on the graph that passes through (0, -2) and whose slope is (1/3).
The line represented by y = -2x + 2 has a y-intercept of 2 and a slope of -2. Draw a line on the graph that passes through (0, 2) and whose slope is -2.
The two lines intersect at a point on the graph. This point represents the solution to the system of equations.
2. Exact Solution using Substitution Method:
Using the first equation, we can solve for x in terms of y:
x - 3y = 6
x = 3y + 6
Now, substitute this expression for x in the second equation:
2x + y = 2
2(3y + 6) + y = 2
6y + 12 + y = 2
7y + 12 = 2
7y = 2 - 12
7y = -10
y = -10/7
Substitute this value of y back into x = 3y + 6:
x = 3(-10/7) + 6
x = -30/7 + 42/7
x = 12/7
So the exact solution to the system of equations is x = 12/7 and y = -10/7.
3. Exact Solution using Elimination Method:
To solve by elimination, we want to eliminate one variable by multiplying one or both equations by a suitable number so that the coefficients of one of the variables will cancel out when the equations are added or subtracted.
Multiply the first equation by 2 and the second equation by -3:
2(x - 3y) = 2(6) -> 2x - 6y = 12
-3(2x + y) = -3(2) -> -6x - 3y = -6
Now, add the two equations together:
(2x - 6y) + (-6x - 3y) = 12 + (-6)
-4x - 9y = 6
Now, solve this new equation for one variable. Let's choose to solve for x:
-4x - 9y = 6
-4x = 6 + 9y
x = (9y + 6)/(-4)
Substitute this expression for x in the first equation:
x - 3y = 6
[(9y + 6)/(-4)] - 3y = 6
(9y + 6) - 12y = 24
9y - 12y = 24 - 6
-3y = 18
y = 18/(-3)
y = -6
Substitute this value of y back into x = (9y + 6)/(-4):
x = (9(-6) + 6)/(-4)
x = (-54 + 6)/(-4)
x = -48/(-4)
x = 12
So the exact solution to the system of equations is x = 12 and y = -6.
Both the substitution and elimination methods yield the same exact solution to the system of equations.