Solve By Addition

Return to Roadmap 2

Start by solving by addition Example 2 from the Chapter 2 Roadmap page. Start with the initial system of two linear equations in two variables.

1x – 4y	=	20	(1)
3x + 4y	=	12	(2)

Now, it conveniently turns out that if we add the two equations, left side of (1) to left side of (2) and right side of (1) to right side of (2), the answer that we get will have eliminated the y variable.

(1x – 4y) + (3x + 4y)	=	20 + 12	(3)
4x	=	32	(4)
x	=	8	(5)

Now that we know the value of x, namely, – 8, we substitute that value into either one of the original equations. We will choose the second.

3(8) + 4y	=	12	(6)
32 + 4y	=	12	(7)
4y	=	^–12	(8)
y	=	^–3	(9)

So our solution is the point where x=8 and y=^–3, that is, the point (8,^–3).

The problem shown above was facilitated by the fact that the coefficients of one of the variables, namely, y, were additive inverses (that is, they add to zero). Here is another problem, but one where the we are not blessed with such a situation.

Solve:

3x + 5y	=	47	(10)
4x + 11y	=	93	(11)

We need to convert these equations into some convient, though equivalent, form so that we produce the same kind of situation that we found in the problem above. If we multiply both sides of the equation (10) by 4 and we multiply both sides of equation (11) by ^–3 we create such a situation.

4(3x + 5y)	=	4(47)	(12)
(-3)(4x + 11y)	=	(-3)(93)	(13)

Or, simplifying:

12x + 20y	=	188	(14)
-12x – 33y	=	-279	(15)

Now we can add the two equations to get (16) and then we can simplify that.

(12x + 20y) + (-12x – 33y)	=	188+(^–279)	(16)
^–13y	=	^–91	(17)
y	=	7	(18)

Now that we know the value of y, namely, 7, we substitute that value into either one of the original equations. In this case we will use equation (10).

3x + 5(7)	=	47	(19)
3x + 35	=	47	(20)
3x	=	12	(21)
x	=	4	(22)

So our solution is the point where x=4 and y=7, that is, the point (4,7).

Return to Roadmap 2
©Roger M. Palay Saline, MI 48176 January, 2017