1. **Stating the problem:** You want to solve a linear system where none of the variables have a coefficient of 1. 2. **General approach:** The goal is to isolate variables or eliminate them to find their values. When coefficients are not 1, you can use methods like substitution, elimination, or matrix operations. 3. **Key formula:** For a system like $$a_1x + b_1y = c_1$$ $$a_2x + b_2y = c_2$$ you can use elimination by multiplying equations to get matching coefficients for one variable. 4. **Example:** Suppose $$3x + 4y = 10$$ $$5x + 2y = 8$$ To eliminate $x$, multiply the first equation by 5 and the second by 3: $$5(3x + 4y) = 5(10) \Rightarrow 15x + 20y = 50$$ $$3(5x + 2y) = 3(8) \Rightarrow 15x + 6y = 24$$ 5. **Subtract equations:** $$ (15x + 20y) - (15x + 6y) = 50 - 24$$ $$ \cancel{15x} + 20y - \cancel{15x} - 6y = 26$$ $$14y = 26$$ 6. **Solve for $y$:** $$y = \frac{26}{14} = \frac{13}{7}$$ 7. **Substitute back to find $x$:** Use the first equation: $$3x + 4\left(\frac{13}{7}\right) = 10$$ $$3x + \frac{52}{7} = 10$$ $$3x = 10 - \frac{52}{7} = \frac{70}{7} - \frac{52}{7} = \frac{18}{7}$$ $$x = \frac{18}{7} \times \frac{1}{3} = \frac{18}{21} = \frac{6}{7}$$ **Summary:** When coefficients are not 1, multiply equations to create matching coefficients, then add or subtract to eliminate variables. Solve for one variable, then substitute back to find others.