For substitution, you would substitute y=3x-4 into 6x + y = 2 ---> 6x + (3x-4) = 2.
6x + 3x - 4 = 2 ---> 9x - 4 =2 ---> 9x = 6 ---> x = 6/9 = 2/3. Then substitute x = 2/3 into one of the original: y = 3(2/3) - 4 = 2-4 = -2. So you have (2/3,-2). Check in the other equation: 6(2/3) + -2 = 2 ---> 4 + -2 = 2 [this checks]
You cant solve it unless it is an equation. To be an equation it must have an equals sign.
-2
This is not Calculus.y=7(Already solved)substiute y=7 into y=8xtherefore 7 = 8xtherefore x = 7/8
isolate
Since the second equation is already solved for "y", you can replace "y" by "9" in the other equation. Then solve the new equation for "x".
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.
You solve this by theoretically diverting the hypotenuse of the x divided by the overall beneficial procedure of y
You cant solve it unless it is an equation. To be an equation it must have an equals sign.
-2
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.
G-Given U-Unknown E-Equation S-Substitution S-Solve
isolate
8840-026
Since the second equation is already solved for "y", you can replace "y" by "9" in the other equation. Then solve the new equation for "x".
You'd need another equation to sub in
the answer