y=2-x 2. Without graphing determine whether each system has one solution , no solution, or infinitely many solutions. a) y=-5x-8 x+3y=4 e) 2x+y=4 2y=-4x+1 b) x-5y=3 2x-10y=-3 c) 3x+2y=-20 d) y=4-3x 6x+2y=8 3x+y=4 f) 6x-8=-2y g) 3x-2y=4 2x-3y=6

## Step 1: Analyze System a)<br />### The first equation is already solved for $y$. Substitute $y = -5x - 8$ into the second equation: $x + 3(-5x - 8) = 4$. Simplify to $x - 15x - 24 = 4$, which becomes $-14x = 28$. Thus, $x = -2$. Substitute $x = -2$ into $y = -5x - 8$ to get $y = -5(-2) - 8 = 10 - 8 = 2$. So, the solution is $(-2, 2)$, and the system has one solution.<br /><br />## Step 2: Analyze System b)<br />### Multiply the first equation by 2: $2(x - 5y) = 2(3)$, which simplifies to $2x - 10y = 6$. The second equation is $2x - 10y = -3$. Since the left sides are the same but the right sides are different, the system has no solution.<br /><br />## Step 3: Analyze System c)<br />### Multiply the second equation by -1: $-1(6x + 2y) = -1(8)$, which simplifies to $-6x - 2y = -8$. Add this to the first equation: $3x + 2y + (-6x - 2y) = -20 + (-8)$. This simplifies to $-3x = -28$, so $x = \frac{28}{3}$. Substitute this into the first equation: $3(\frac{28}{3}) + 2y = -20$, which simplifies to $28 + 2y = -20$. Thus, $2y = -48$, and $y = -24$. The system has one solution $(\frac{28}{3}, -24)$.<br /><br />## Step 4: Analyze System d)<br />### Rewrite the first equation as $3x + y = 4$. This is the same as the second equation. Since both equations are the same, the system has infinitely many solutions.<br /><br />## Step 5: Analyze System e)<br />### Rewrite the second equation as $4x + 2y = 1$. Multiply the first equation by 2: $2(2x + y) = 2(4)$, which simplifies to $4x + 2y = 8$. Since the left sides are the same but the right sides are different, the system has no solution.<br /><br />## Step 6: Analyze System f)<br />### Rewrite the first equation as $6x + 2y = 8$. Divide by 2: $3x + y = 4$. The second equation can be rewritten as $2x + 3y = 6$. Multiply the first equation by 3: $9x + 3y = 12$. Subtract the second equation from this: $(9x + 3y) - (2x + 3y) = 12 - 6$, which simplifies to $7x = 6$, so $x = \frac{6}{7}$. Substitute this into $3x + y = 4$: $3(\frac{6}{7}) + y = 4$, which means $\frac{18}{7} + y = 4$. Thus, $y = 4 - \frac{18}{7} = \frac{28 - 18}{7} = \frac{10}{7}$. The system has one solution $(\frac{6}{7}, \frac{10}{7})$.<br /><br />## Step 7: Analyze System g)<br />### This system is already in standard form. We can use any method to solve. Multiply the first equation by 3 and the second equation by -2: $9x - 6y = 12$ and $-4x + 6y = -12$. Add these equations: $5x = 0$, so $x = 0$. Substitute $x = 0$ into the first equation: $3(0) - 2y = 4$, so $-2y = 4$, and $y = -2$. The system has one solution $(0, -2)$.