Question

solving systems of linear equations: substitution
instruction active
identifying a system with no solution
which system has no solution?

y = -3x + 8
6x + 2y = -4.5

y = 9x + 6.25
-18x + 2y = 12.5

y = 4.5x - 5
-3x + 2y = 6

y = 3x + 9
x + 8y = 12.3

Explanation:

Step1: Recall no-solution condition

A system of linear equations $a_1x+b_1y=c_1$ and $a_2x+b_2y=c_2$ has no solution if $\frac{a_1}{a_2}=\frac{b_1}{b_2}
eq\frac{c_1}{c_2}$.

Step2: Analyze first system

Rewrite $y=-3x+8$ as $3x+y=8$.
Check ratios: $\frac{3}{6}=\frac{1}{2}$, $\frac{8}{-4.5}=-\frac{16}{9}$. $\frac{1}{2}
eq-\frac{16}{9}$, so has solution.

Step3: Analyze second system

Rewrite $y=9x+6.25$ as $9x-y=-6.25$.
Check ratios: $\frac{9}{-18}=-\frac{1}{2}$, $\frac{-1}{2}=-\frac{1}{2}$, $\frac{-6.25}{12.5}=-\frac{1}{2}$. All ratios equal, so has infinite solutions.

Step4: Analyze third system

Rewrite $y=4.5x-5$ as $4.5x-y=5$.
Check ratios: $\frac{4.5}{-3}=-\frac{3}{2}$, $\frac{-1}{2}=-\frac{1}{2}$. $-\frac{3}{2}
eq-\frac{1}{2}$, so has solution.

Step5: Analyze fourth system

Rewrite $y=3x+9$ as $3x-y=-9$.
Check ratios: $\frac{3}{1}=3$, $\frac{-1}{8}=-\frac{1}{8}$. $3
eq-\frac{1}{8}$, so has solution.
Correction: Recheck first system formatting
Rewrite $y=-3x+8$ as $3x + y = 8$, multiply by 2: $6x+2y=16$. Compare to $6x+2y=-4.5$.
Now $\frac{6}{6}=\frac{2}{2}=1$, $\frac{16}{-4.5}
eq1$. This satisfies no-solution condition.

Answer:

The system with no solution is:
$y = -3x + 8$
$6x + 2y = -4.5$