Solving a Linear System by Substitution
This first video shows an example on how to solve a system of linear equations using substitution when one of the equations already has an isolated variable. That is the first thing to check for! Make sure one equation has an isolated variable, then you can substitute the expression from that equation into the other equation.
The second video demonstrates what to do when you need to solve a system of equations but neither of the two equations has an isolated variable. When selecting the equation to rewrite, select the equation that has a variable term that has a coefficient of positive 1, and then solve for that variable. For example, if the system consists of the equations \( 2x + y = 8 \) and \( 3x + 5y = 10 \), rewriting the first equation \( 2x + y = 8 \) would be more efficient. Since the "\( y \)" has the coefficient of positive 1, you would solve for the y, rewriting that equation as \( y = -2x + 8 \). If the system consisted of the equations \( 5x - 2y = 9 \) and \( x + 3y = 5 \), rewriting the second equation \(x + 3y = 5 \) would be more efficient. Since the "\( x \)" has the coefficient of positive 1, you would solve for \( x \), making the equation \( x = -3y + 5 \). Using this idea of looking for the variable term with a coefficient of 1, will help to make substitution more smooth.