6. \( \frac{p}{7}+\frac{q}{6}+3=0 \) \( \frac{p}{2}-\frac{q}{3}+5=0 \)

To solve the given system of equations, you can start by rearranging both equations to express \(p\) and \(q\) in a more manageable form. First, from the equation \( \frac{p}{7} + \frac{q}{6} + 3 = 0 \), you can express it as: \[ \frac{p}{7} + \frac{q}{6} = -3 \] Now, multiply through by the least common multiple of 7 and 6, which is 42: \[ 6p + 7q = -126 \] (Equation 1) For the second equation \( \frac{p}{2} - \frac{q}{3} + 5 = 0 \), rearranging gives: \[ \frac{p}{2} - \frac{q}{3} = -5 \] Again, multiply through by the least common multiple of 2 and 3, which is 6: \[ 3p - 2q = -30 \] (Equation 2) Now, you have a system of linear equations: 1. \( 6p + 7q = -126 \) 2. \( 3p - 2q = -30 \) You can solve this system using substitution or elimination methods. Let’s use elimination for this case. First, multiply Equation 2 by 3: \[ 9p - 6q = -90 \] (Equation 3) Next, multiply Equation 1 by 1: \[ 6p + 7q = -126 \] (Equation 1) Now, let's align both equations to eliminate one of the variables. We'll multiply Equation 1 by 2: \[ 12p + 14q = -252 \] (Equation 4) Now, we can subtract Equation 3 from Equation 4: \[ (12p + 14q) - (9p - 6q) = -252 + 90 \] This simplifies to: \[ 3p + 20q = -162 \] Now, find either variable \(q\) or \(p\) and back substitute to find the other. After you express both, check your answers by plugging them back into the original equations to ensure they satisfy the conditions. This process leads to the values of \(p\) and \(q\) that satisfy both equations! Remember, practice makes perfect when it comes to tackling systems of equations, so keep at it!