Geometry & Figures

Depth · one concept from first principles · step 2 of 4

A zero product graphs as two lines

Count the points (x, y) that satisfy both (y-2)(x-2y+2)=0 and (2x+y-5)(x-2y-2)=0.
Expressions
Your answer
How to solve
Strategy Organize Information in More Ways — Re-read each equation: a product is zero when one factor is zero, so each equation's graph is a PAIR of lines. A common solution is a crossing between a line from the first pair and a line from the second, so list the four pairings and check which actually meet.
1STEP 1

Split each equation into two lines

Equation 1 is y=2 or y=(1/2)x+1; equation 2 is y=-2x+5 or y=(1/2)x-1.

2STEP 2

Pair 1 with 1, and 1 with 2

y=2 with y=-2x+5: 2=-2x+5, so (3/2, 2). y=2 with y=(1/2)x-1: 2=(1/2)x-1, so (6, 2).

3STEP 3

The other two pairs

y=(1/2)x+1 meets y=-2x+5 at (8/5, 9/5); it is parallel to y=(1/2)x-1 (same slope), so that pair never meets.

4STEP 4

Count distinct points

The three points (3/2, 2), (6, 2), (8/5, 9/5) are all different, so there are 3 ordered pairs.

1+1+1+0=3
Answer
3
Plug in (8/5, 9/5): 8/5-18/5+2=0 makes equation 1 zero, and 16/5+9/5-5=0 makes equation 2 zero. The other two points sit on y=2 and on one line of equation 2 by construction.
Takeaway

A product equal to zero is really two simpler equations wearing one coat.

  • Split each equation into two lines
  • Pair 1 with 1, and 1 with 2
  • The other two pairs
  • Count distinct points
Where next?
Another one like thissuggested

A parent dashboard for the family lives at sensimlab.com.

▶ Practice — 12 problems