Systems of equations

SAT systems reward noticing structure. Before solving, compare the equations. They may be multiples of each other, or adding them may hand you the answer directly.

If...	Then
Different ratios $\frac{a}{d} \neq = \frac{b}{e}$	One solution
$\frac{a}{d} = \frac{b}{e} \neq = \frac{c}{f}$	No solution
$\frac{a}{d} = \frac{b}{e} = \frac{c}{f}$	Infinitely many

Work through it

$2 x + 3 y = 6$ and $4 x + k y = 10$ has no solution. Find $k$ .

No solution means the left sides are proportional. The $x$ coefficient doubled. $k = 6$

Check the right side: 6 doubled is 12, not 10.no solution holds

$3 x + y = 14$ and $x + 3 y = 10$ . Find $x + y$ .

Don't solve. Add the equations. $4 x + 4 y = 24$

Divide by 4. $x + y = 6$

Tickets: 50 sold, adults $8, kids $5, total $310. How many adults?

Two equations. $a + k = 50$ , $8 a + 5 k = 310$

Substitute $k = 50 - a$ . $8 a + 250 - 5 a = 310$

Solve. $a = 20$

Watch out for this

Finding $x$ when the question asks for $x + y$ .

Read the last line first. Often you don't need each value at all.

The short version
Same slope and different intercepts means no solution. Equations that are multiples of each other mean infinitely many.

Your turn

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