The 'substitution method' is used when you have two equations in two variables.
It is only practical if one of the equations is set up so that you can easily solve for one of the variables in terms of the other.
You would then "substitute" that into the other equation.
In your case you only gave one equation.
Did you want to solve for x??
if so, then multiply each term by 49 to eliminate the fractions
49x + 42 = 294x + 78
-36 = 245x
x = -36/245
Since you did not supply any brackets to change the order of operation, that would be the correct answer the way you wrote the question.
x + 6 / 7 = 6x + 78 / 49
My mind is a bit rusty and I can't remember how to use this method
2 answers
Example:
2y + x = 3
(1)
4y – 3x = 1
(2)
Equation 1 looks like it would be easy to solve for x, so we take it and isolate x:
2y + x = 3
x = 3 – 2y
(3)
Now we can use this result and substitute 3 - 2y in for x in equation 2:
Now that we have y, we still need to substitute back in to get x. We could substitute back into any of the previous equations, but notice that equation 3 is already conveniently solved for x:
And so the solution is (1, 1).
As a rule, the substitution method is easier and quicker than the addition method when one of the equations is very simple and can readily be solved for one of the variables.
2y + x = 3
(1)
4y – 3x = 1
(2)
Equation 1 looks like it would be easy to solve for x, so we take it and isolate x:
2y + x = 3
x = 3 – 2y
(3)
Now we can use this result and substitute 3 - 2y in for x in equation 2:
Now that we have y, we still need to substitute back in to get x. We could substitute back into any of the previous equations, but notice that equation 3 is already conveniently solved for x:
And so the solution is (1, 1).
As a rule, the substitution method is easier and quicker than the addition method when one of the equations is very simple and can readily be solved for one of the variables.
7 answers