Linear Elimination F48B02

1. **State the problem:** We need to solve the simultaneous linear equations: $$4a + 3b = -6$$ $$5a + 5b + 7 = 0$$ 2. **Rewrite the second equation:** Move the constant term to the right side: $$5a + 5b = -7$$ 3. **Use the elimination method:** To eliminate one variable, multiply the first equation by 5 and the second by 4 to align coefficients of $a$: $$5 \times (4a + 3b) = 5 \times (-6) \Rightarrow 20a + 15b = -30$$ $$4 \times (5a + 5b) = 4 \times (-7) \Rightarrow 20a + 20b = -28$$ 4. **Subtract the first new equation from the second:** $$ (20a + 20b) - (20a + 15b) = -28 - (-30) $$ $$ 20a - 20a + 20b - 15b = -28 + 30 $$ $$ 5b = 2 $$ 5. **Solve for $b$:** $$ b = \frac{2}{5} $$ 6. **Substitute $b$ back into one original equation to find $a$:** Use the first equation: $$ 4a + 3 \times \frac{2}{5} = -6 $$ $$ 4a + \frac{6}{5} = -6 $$ 7. **Isolate $a$:** $$ 4a = -6 - \frac{6}{5} = -\frac{30}{5} - \frac{6}{5} = -\frac{36}{5} $$ $$ a = \frac{-\frac{36}{5}}{4} = -\frac{36}{5} \times \frac{1}{4} = -\frac{36}{20} $$ 8. **Simplify $a$:** $$ a = -\frac{9}{5} $$ **Final answer:** $$ a = -\frac{9}{5}, \quad b = \frac{2}{5} $$