How did you get: x = 2; y = -1
for
Directions: Sove the system by addition
5x-3y=13
4x-3y=11
for this type of problem there is no shading correct. or am i wrong
5x-3y=13
4x-3y=11
If you subtract the two equations you get:
5x-3y - (4x-3y) = 13 - 11
The y's on the left hand side cancel:
5x - 4x - 3y + 3y = 13 - 11 --->
x = 2
If you insert x = 2 in either of the two original equations and soleve for y you find that y = -1.
To solve the system of equations using the method of addition, follow these steps:
1. Write down both equations:
5x - 3y = 13
4x - 3y = 11
2. In order to eliminate one variable, multiply one equation (or both) by a constant so that the coefficients of one of the variables are the same in both equations. In this case, the coefficients of 'y' in both equations are already the same, so we can proceed.
3. Subtract one equation from the other to cancel out the variable. In this case, subtract the second equation from the first equation:
(5x - 3y) - (4x - 3y) = 13 - 11
4. Simplify the equation:
5x - 4x - 3y + 3y = 13 - 11
x = 2
5. Now that you have the value of 'x', substitute it back into either of the original equations to solve for 'y'. Let's substitute 'x = 2' into the first equation:
5(2) - 3y = 13
10 - 3y = 13
6. Solve for 'y':
-3y = 13 - 10
-3y = 3
y = -1
So, the solution for the system of equations is:
x = 2
y = -1
Regarding your shading question, shading is typically used when graphing inequalities to represent the solution region. In this case, since you have two equations and not inequalities, there is no need for shading. Shading is used to show the solution region in graphs, not when solving systems algebraically.