Find the value of k which makes the following simultaneous equations to have no solution: y/5 - 1 = x/15 k/40 * x - y/8 = 1

Steps to Solve

1. Rewrite the first equation Multiply both sides of the first equation by $15$ to eliminate the fractions: $$15(\frac{y}{5} - 1) = 15(\frac{x}{15})$$ $$3y - 15 = x$$ $$x = 3y - 15$$

2. Rewrite the second equation Multiply both sides of the second equation by $40$ to eliminate the fractions: $$40(\frac{k}{40}x - \frac{y}{8}) = 40(1)$$ $$kx - 5y = 40$$

3. Condition for no solution For a system of linear equations to have no solution, the ratio of the coefficients of $x$ and $y$ must be equal, but the ratio of the constants must be different. First we can rewrite equations in the form $ax + by = c$: Equation 1: $x - 3y = -15$ Equation 2: $kx - 5y = 40$ For no solution, we need: $$ \frac{a_1}{a_2} = \frac{b_1}{b_2} \neq \frac{c_1}{c_2}$$ Therefore $$ \frac{1}{k} = \frac{-3}{-5} \neq \frac{-15}{40}$$

4. Solve for k Solve $\frac{1}{k} = \frac{3}{5}$ for $k$: $$3k = 5$$ $$k = \frac{5}{3}$$

5. Verify the inequality condition Check if $\frac{3}{5} \neq \frac{-15}{40}$: $$\frac{3}{5} \neq \frac{-3}{8}$$ Since $\frac{3}{5}$ is not equal to $\frac{-3}{8}$, the condition for no solution is satisfied.