Find p when a system of linear equations has no solution

Key takeaway: For a linear system to have no solution, the lines must be parallel but distinct. That means the coefficients of $x$ and $y$ have to match in slope, while the constant terms must not.

We first rewrite each equation in a clearer form.

Equation 1

$\frac78 y-\frac58 x=\frac47-\frac78 y$ Bring the $y$ terms together: $\frac78 y+\frac78 y=\frac58 x+\frac47$ So $\frac74 y=\frac58 x+\frac47$ Multiply by 4: $7y=\frac52 x+\frac{16}{7}$ Hence $y=\frac{5}{14}x+\frac{16}{49}$

Equation 2

$\frac54 x+\frac74=py+\frac{15}{4}$ Rearrange: $py=\frac54 x-\frac12$ So $y=\frac{5}{4p}x-\frac{1}{2p}$

For no solution, the two lines must be parallel, so their slopes must be equal: $\frac{5}{14}=\frac{5}{4p}$ Cancel 5: $\frac{1}{14}=\frac{1}{4p}$ Thus $4p=14$ $p=\frac72$

Now check the intercepts: they are different, so the lines are distinct. Therefore the system has no solution when $\boxed{\frac72}$

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Pitfalls the pros know 👇 A common mistake is to match the constant terms first. That does not tell you whether a system has no solution. What matters is equal slopes with different intercepts. Also be careful when rearranging the first equation; the $y$ terms need to be collected correctly before simplifying.

What if the problem changes? If the question had asked for infinitely many solutions, then the two equations would need to represent exactly the same line, so both the slope and intercept would have to match. That would give an extra condition on the constants, not just on $p$.

Tags: parallel lines, consistent system, slope-intercept form