College Algebra, 2013a

8.1 Systems of Linear Equations: Gaussian Elimination 551
4
y
2
y
(2, 3)
2
1
(−2, 1)
1
−4 −3 −2 −1
−1
x
−1 1 2 3 4
x
−2
2x − y =1
y =3
3x +4y = −2
−3x − y =5
3. The equations in the third system are more approachable if we clear denominators. We
multiply both sides of the first equation by 15 and both sides of the second equation by 18
to obtain the kinder, gentler system
{ 5x − 12y = 21
4x +6y = 9
Adding these two equations directly fails to eliminate either of the variables, but we note
that if we multiply the first equation by 4 and the second by −5, we will be in a position to
eliminate the x term
20x − 48y = 84
+ (−20x − 30y = −45)
−78y = 39
From this we get y = − 1 2
. We can temporarily avoid too much unpleasantness by choosing to
substitute y = − 1 2
into one of the equivalent equations we found by clearing denominators,
say into 5x − 12y = 21. We get 5x + 6 = 21 which gives x = 3. Our answer is ( 3, − 1 2)
.
At this point, we have no choice − in order to check an answer algebraically, we must see
if the answer satisfies both of the original equations, so we substitute x = 3 and y = − 1 2
into both x 3 − 4y 5 = 7 5 and 2x 9 + y 3 = 1 2
. We leave it to the reader to verify that the solution
is correct. Graphing both of the lines involved with considerable care yields an intersection
point of ( 3, − 1 2)
.
4. An eerie calm settles over us as we cautiously approach our fourth system. Do its friendly
integer coefficients belie something more sinister? We note that if we multiply both sides of
the first equation by 3 and both sides of the second equation by −2, we are ready to eliminate
the x