Solve by the elimination process:
0.3x - 0.2y = 4;
0.2x + 0.3y = 5/23
multiply the first by .3, the second equation by .2, then add the equations.
thank you so much
To solve the given system of equations using the elimination process, you can eliminate one of the variables by adding or subtracting the two equations.
In this case, let's eliminate the y variable. To do so, we need to manipulate the two equations such that the coefficients of the y terms will cancel each other out when added or subtracted together.
First, multiply the first equation by 10 to get rid of the decimal points:
10(0.3x - 0.2y) = 10(4)
3x - 2y = 40
Now, multiply the second equation by 10 to get rid of the decimal points:
10(0.2x + 0.3y) = 10(5/23)
2x + 3y = 50/23
Now, we have the following system of equations:
3x - 2y = 40
2x + 3y = 50/23
Next, we need to eliminate the y variable by adding the two equations together. To do this, we can multiply the first equation by 3 and the second equation by 2 to make the coefficients of the x terms cancel out:
3(3x - 2y) + 2(2x + 3y) = 3(40) + 2(50/23)
9x - 6y + 4x + 6y = 120 + 100/23
13x = 2760/23 + 100/23
13x = 2860/23
x = (2860/23) / 13
x = 220/23
Now substitute the value of x into one of the original equations. Let's use the first equation:
0.3x - 0.2y = 4
0.3(220/23) - 0.2y = 4
66/23 - 0.2y = 4
-0.2y = 4 - 66/23
-0.2y = 92/23 - 66/23
-0.2y = 26/23
y = (26/23) / (-0.2)
y = -26/46
y = -13/23
Therefore, the solution to the system of equations is x = 220/23 and y = -13/23.