|French||A Level||£24 /hr|
Dan (Student) February 7 2017
Dan (Parent) January 31 2017
Dan (Student) January 31 2017
Dan (Parent) January 24 2017
Always follow the same three steps : equating – substracting – substituting but you can have different cases :
Case 1: we only need to multiply one equation
Note : Always check first that the equation is solvable (= there are as many or more equations than there are unknowns.)
We notice that 6x is a multiple of 2x. So we only need to multiply the second equation to equate the coefficients.
Substracting the 2nd equation from 3 times the first one gives 19y=19 so y=1
Now we substitute into the first equation to get
The solution is x=2, y=1
Case 2 : we need to multiply both equations
3x+5y = 14
We proceed the same way as in the first case but to equate coefficients we need to multiply both equations, because the coefficients are not multiples of each other.
Multiplying the first equation by 3 and substracting 5 times the second gives 11x=33 so x=3.
Now we substitute back in to get
The solution is x=3, y=1see more