Linear Systems Elimination 5B1250

1. **Problem Statement:** Solve each linear system using the method of elimination. --- ### a) System: $$\ell_1: 3x + 4y = 4$$ $$\ell_2: -2x - 4y = 8$$ **Step 1:** Add the two equations to eliminate $y$: $$ (3x + 4y) + (-2x - 4y) = 4 + 8 $$ $$ 3x - 2x + 4y - 4y = 12 $$ $$ x + \cancel{4y} - \cancel{4y} = 12 $$ $$ x = 12 $$ **Step 2:** Substitute $x=12$ into $\ell_1$: $$ 3(12) + 4y = 4 $$ $$ 36 + 4y = 4 $$ $$ 4y = 4 - 36 $$ $$ 4y = -32 $$ $$ y = \frac{-32}{4} = -8 $$ **Solution for (a):** $$ x = 12, \quad y = -8 $$ --- ### b) System: $$\ell_1: 3x + 5y = 1$$ $$\ell_2: 4x + y = 7$$ **Step 1:** Multiply $\ell_2$ by $-5$ to align $y$ coefficients: $$ -5(4x + y) = -5(7) $$ $$ -20x - 5y = -35 $$ **Step 2:** Add to $\ell_1$: $$ (3x + 5y) + (-20x - 5y) = 1 + (-35) $$ $$ 3x - 20x + 5y - 5y = -34 $$ $$ -17x + \cancel{5y} - \cancel{5y} = -34 $$ $$ -17x = -34 $$ $$ x = \frac{-34}{-17} = 2 $$ **Step 3:** Substitute $x=2$ into $\ell_2$: $$ 4(2) + y = 7 $$ $$ 8 + y = 7 $$ $$ y = 7 - 8 = -1 $$ **Solution for (b):** $$ x = 2, \quad y = -1 $$ --- ### c) System: $$\ell_1: 2x - 5y = 2$$ $$\ell_2: -5x = -3y + 2$$ Rewrite $\ell_2$ to standard form: $$ -5x + 3y = 2 $$ **Step 1:** Multiply $\ell_1$ by $3$ and $\ell_2$ by $5$ to align $y$ coefficients: $$ 3(2x - 5y) = 3(2) \Rightarrow 6x - 15y = 6 $$ $$ 5(-5x + 3y) = 5(2) \Rightarrow -25x + 15y = 10 $$ **Step 2:** Add the two equations: $$ (6x - 15y) + (-25x + 15y) = 6 + 10 $$ $$ 6x - 25x - 15y + 15y = 16 $$ $$ -19x + \cancel{-15y} + \cancel{15y} = 16 $$ $$ -19x = 16 $$ $$ x = \frac{16}{-19} = -\frac{16}{19} $$ **Step 3:** Substitute $x = -\frac{16}{19}$ into $\ell_1$: $$ 2\left(-\frac{16}{19}\right) - 5y = 2 $$ $$ -\frac{32}{19} - 5y = 2 $$ $$ -5y = 2 + \frac{32}{19} $$ $$ -5y = \frac{38}{19} + \frac{32}{19} = \frac{70}{19} $$ $$ y = -\frac{70}{19} \times \frac{1}{5} = -\frac{70}{95} = -\frac{14}{19} $$ **Solution for (c):** $$ x = -\frac{16}{19}, \quad y = -\frac{14}{19} $$