Question 397096: How do u work this problem out? Its system of equations
3x+4y=12
Y=-2x+10
Answer by Sarpi(32)   (Show Source):
You can put this solution on YOUR website! Solution
This is a kind of two simultaneous equation.
Step 1:
Arrange the equation as follows;
3x+4y = 12 -------------- eqn 1
y = 10-2x ----------- eqn 2
(I just interchanged the position of eqn 2 but it is the same as y = -2x+10)
Step 2:
By the substitution method, you make one of the variable x or y the subject in one of the eqn.
Luckily, we have y has the subject in eqn 2.
ie y = 10 - 2x ---------- eqn 3.
Step 3:
You substitute or put the eqn 3 into eqn 1. Note you cannot put eqn 3 into its derived eqn (ie eqn 2).
=> where y should be replaced by 10-2x in eqn 1
ie in eqn 1;
3x+4y = 12
3x+4(10-2x) = 12 "10-2x is put in bracket because it is multiplying by 4"
3x+40-8x = 12 "multiplication; 4*10 =40 and 4*-2x=-8x"
3x-8x = 12- 40 "grouping of liked variables(numbers with x on one side"
-5x = -28
x = -28/-5 "we obtain this by dividing both side by -5"
x = 28/5 or 5.6
Step 4:
Now that we have found x, we move on to find y using one of the eqns(1 or 2). Thus it is easy to use eqn 2 or 3 in order to find y.
=> y = 10-2x
y = 10-2(5.6)
y = 10-11.2
y = -1.2
Hence the solution to the problem is x=5.6 and y=-1.2
By checking;
3x+4y=12
3(5.6)+4(-1.2) will exactly give you 12.
And I hope you will easily understand my steps. Cheers!
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