1. **State the problem:** Find the points of intersection for each system of linear equations given. 2. **Recall the method:** The point of intersection of two lines $y=m_1x+b_1$ and $y=m_2x+b_2$ is found by setting the equations equal: $$m_1x+b_1 = m_2x+b_2$$ and solving for $x$, then substituting back to find $y$. 3. **Solve each system:** **System 1:** $$y=2x+1$$ $$y=x+3$$ Set equal: $$2x+1 = x+3$$ Subtract $x$ both sides: $$\cancel{2x}+1 = \cancel{x}+3 \Rightarrow x+1=3$$ Subtract 1: $$x=2$$ Find $y$: $$y=2(2)+1=4+1=5$$ Point: $(2,5)$ **System 2:** $$y=x+3$$ $$y=-3x-2$$ Set equal: $$x+3 = -3x-2$$ Add $3x$: $$x+3x+3 = -3x+3x-2 \Rightarrow 4x+3 = -2$$ Subtract 3: $$4x = -5$$ Divide by 4: $$\frac{4x}{\cancel{4}} = \frac{-5}{\cancel{4}} \Rightarrow x = -\frac{5}{4}$$ Find $y$: $$y = -\frac{5}{4} + 3 = -\frac{5}{4} + \frac{12}{4} = \frac{7}{4}$$ Point: $\left(-\frac{5}{4}, \frac{7}{4}\right)$ **System 3:** $$y = -x - 7$$ $$y = x + 3$$ Set equal: $$-x - 7 = x + 3$$ Add $x$: $$-x + x - 7 = x + x + 3 \Rightarrow -7 = 2x + 3$$ Subtract 3: $$-10 = 2x$$ Divide by 2: $$\frac{-10}{\cancel{2}} = \frac{2x}{\cancel{2}} \Rightarrow x = -5$$ Find $y$: $$y = -(-5) - 7 = 5 - 7 = -2$$ Point: $(-5, -2)$ **System 4:** $$y = -x - 7$$ $$y = 2x + 1$$ Set equal: $$-x - 7 = 2x + 1$$ Add $x$: $$-x + x - 7 = 2x + x + 1 \Rightarrow -7 = 3x + 1$$ Subtract 1: $$-8 = 3x$$ Divide by 3: $$\frac{-8}{\cancel{3}} = \frac{3x}{\cancel{3}} \Rightarrow x = -\frac{8}{3}$$ Find $y$: $$y = 2\left(-\frac{8}{3}\right) + 1 = -\frac{16}{3} + 1 = -\frac{16}{3} + \frac{3}{3} = -\frac{13}{3}$$ Point: $\left(-\frac{8}{3}, -\frac{13}{3}\right)$ **System 5:** $$y = 2x + 1$$ $$y = -3x - 2$$ Set equal: $$2x + 1 = -3x - 2$$ Add $3x$: $$2x + 3x + 1 = -3x + 3x - 2 \Rightarrow 5x + 1 = -2$$ Subtract 1: $$5x = -3$$ Divide by 5: $$\frac{5x}{\cancel{5}} = \frac{-3}{\cancel{5}} \Rightarrow x = -\frac{3}{5}$$ Find $y$: $$y = 2\left(-\frac{3}{5}\right) + 1 = -\frac{6}{5} + 1 = -\frac{6}{5} + \frac{5}{5} = -\frac{1}{5}$$ Point: $\left(-\frac{3}{5}, -\frac{1}{5}\right)$ 4. **Final answers:** - System 1: $(2,5)$ - System 2: $\left(-\frac{5}{4}, \frac{7}{4}\right)$ - System 3: $(-5,-2)$ - System 4: $\left(-\frac{8}{3}, -\frac{13}{3}\right)$ - System 5: $\left(-\frac{3}{5}, -\frac{1}{5}\right)$