Question

System practice problems

Answer 1

Looking at the first system of equations,

16x - 10y = 10

-8x - 6y = 6

If we multiply both sides of the second equation by 2, the coefficient of x is exactly the negative of the coefficient of x in the first equation.

-8x - 6y = 6

⇒   2 (-8x - 6y) = 2 (6)

⇒   -16x - 12y = 12

By combining this new equation with the first one, we can eliminate x and solve for y :

(16x - 10y) + (-16x - 12y) = 10 + 12

⇒   -22y = 22

⇒   y = -1

Then we just solve for x by replacing y in either equation.

16x - 10y = 10

⇒   16x - 10 (-1) = 10

⇒   16x + 10 = 10

⇒   16x = 0

⇒   x = 0

The main idea behind elimination is combining the given equations in just the right amount so that one of the variables disappears. The "right amount" involves using the LCM of the coefficients of a given variable. In this example, the x-coefficients had LCM(8, 16) = 16, so we only had to scale one of the equations (the one with -8x) to cancel all the x terms.

If we wanted to eliminate y first instead, we first note that LCM(6, 10) = 30. To get 30 as a coefficient on y, in the first equation we would have multiplied by 3:

16x - 10y = 10

⇒   3 (16x - 10y) = 3 (10)

⇒   48x - 30y = 30

And in the second equation, we would have multiplied by -5 (negative so that upon combining the equations, we end up with -30y + 30y = 0):

-8x - 6y = 6

⇒   -5 (-8x - 6y) = -5 (6)

⇒   40x + 30y = -30

Now combining the two scaled equations gives

(48x - 30y) + (40x + 30y) = 30 + (-30)

⇒   88x = 0

⇒   x = 0

We then solve for y :

16x - 10y = 10

⇒   -10y = 10

⇒   y = -1

so we end up with the same solution as before.