Question

1. $x - 2y = -3$

$y = x + 1$
$x-2(x+1)=-3$

1. $x + y = 3$

$2x + 4y = 8$
$x=12+5$

Explanation:

System 1: $x - 2y = -3$; $y = x + 1$

Step1: Substitute $y=x+1$ into first equation

$x - 2(x + 1) = -3$

Step2: Expand and simplify for $x$

$x - 2x - 2 = -3$
$-x - 2 = -3$
$-x = -3 + 2$
$-x = -1$
$x = 1$

Step3: Substitute $x=1$ into $y=x+1$

$y = 1 + 1 = 2$

---

System 2: $x + y = 3$; $2x + 4y = 8$

Step1: Simplify the second equation

Divide by 2: $x + 2y = 4$

Step2: Subtract first equation from simplified equation

$(x + 2y) - (x + y) = 4 - 3$

Step3: Solve for $y$

$y = 1$

Step4: Substitute $y=1$ into $x+y=3$

$x + 1 = 3$
$x = 3 - 1 = 2$

Answer:

1. For the system

$$\begin{cases} x - 2y = -3 \\ y = x + 1 \end{cases}$$

, the solution is $x=1$, $y=2$

1. For the system

$$\begin{cases} x + y = 3 \\ 2x + 4y = 8 \end{cases}$$

, the solution is $x=2$, $y=1$