Solve the following systems of equations graphically: a) x + y = 3 ; x - y = 1 b) x + 2y = 2 ; 2x + 5y = 2

Steps to Solve

1. Solve for y in terms of x for each equation in system a)

   • For $x + y = 3$, subtract $x$ from both sides to get $y = 3 - x$.
   • For $x - y = 1$, subtract $x$ from both sides to get $-y = 1 - x$, then multiply both sides by $-1$ to get $y = x - 1$.

2. Find two points for each equation in system a)

   • For $y = 3 - x$: If $x = 0$, then $y = 3 - 0 = 3$. Point 1: $(0, 3)$. If $x = 3$, then $y = 3 - 3 = 0$. Point 2: $(3, 0)$.

   • For $y = x - 1$: If $x = 0$, then $y = 0 - 1 = -1$. Point 3: $(0, -1)$. If $x = 1$, then $y = 1 - 1 = 0$. Point 4: $(1, 0)$.

3. Plot the lines for system a) and find the intersection point

   • Plot the points and draw the lines for $y = 3 - x$ and $y = x - 1$. The intersection point of these two lines is the solution to the system. The lines intersect at $(2, 1)$.

4. Solve for y in terms of x for each equation in system b)

• For $x + 2y = 2$, subtract $x$ from both sides to get $2y = 2 - x$, then divide both sides by $2$ to get $y = 1 - \frac{1}{2}x$.
• For $2x + 5y = 2$, subtract $2x$ from both sides to get $5y = 2 - 2x$, then divide both sides by $5$ to get $y = \frac{2}{5} - \frac{2}{5}x$.

5. Find two points for each equation in system b)

   • For $y = 1 - \frac{1}{2}x$: If $x = 0$, then $y = 1 - \frac{1}{2}(0) = 1$. Point 1: $(0, 1)$. If $x = 2$, then $y = 1 - \frac{1}{2}(2) = 0$. Point 2: $(2, 0)$.

   • For $y = \frac{2}{5} - \frac{2}{5}x$: If $x = 1$, then $y = \frac{2}{5} - \frac{2}{5}(1) = 0$. Point 3: $(1,0)$. If $x = -4$, then $y = \frac{2}{5} - \frac{2}{5}(-4) = 2$. Point 4: $(-4, 2)$.

6. Plot the lines for system b) and find the intersection point

   • Plot the points and draw the lines for $y = 1 - \frac{1}{2}x$ and $y = \frac{2}{5} - \frac{2}{5}x$. The intersection point of these two lines is the solution to the system. The lines intersect at $(6, -2)$.