i am guessing here that this is what you mean as it involves substitution:
2x-2y = 4
therefore -2y = 4 - 2x
Sub the above line into the first equation gives:
13x - 2x +4 = 70
11x = 66
x=6
then sub in x
13(6) - 2y = 70
78 - 2y = 70
y=4
Yep, this is the exact same as simultaneous equations
Accuracy.
-2
This is not Calculus.y=7(Already solved)substiute y=7 into y=8xtherefore 7 = 8xtherefore x = 7/8
If you mean: x+7y = 39 and 3x-2y = 2 Then by substitution: x = 4 and y = 5
One way to solve this system of equations is by using matrices. Form an augmented matrix in which the first 2x2 matrix is the coefficient matrix and the 2x1 matrix on its right is the answer. Now apply Gaussian Elimination and back-substitution. Using this method gives x=5 and y=1.
Accuracy.
-2
You can solve lineaar quadratic systems by either the elimination or the substitution methods. You can also solve them using the comparison method. Which method works best depends on which method the person solving them is comfortable with.
This is not Calculus.y=7(Already solved)substiute y=7 into y=8xtherefore 7 = 8xtherefore x = 7/8
There are no disadvantages. There are three main ways to solve linear equations which are: substitution, graphing, and elimination. The method that is most appropriate can be found by looking at the equation.
If you mean: x+7y = 39 and 3x-2y = 2 Then by substitution: x = 4 and y = 5
You solve this by theoretically diverting the hypotenuse of the x divided by the overall beneficial procedure of y
Using the u substitution method of derivation (selecting sinx as u and cosxdx as du), you get f'(x)=cscx.
G-Given U-Unknown E-Equation S-Substitution S-Solve
One way to solve this system of equations is by using matrices. Form an augmented matrix in which the first 2x2 matrix is the coefficient matrix and the 2x1 matrix on its right is the answer. Now apply Gaussian Elimination and back-substitution. Using this method gives x=5 and y=1.
You cant solve it unless it is an equation. To be an equation it must have an equals sign.
x = -2 and y = 1