2 Solving Linear Systems There are two methods of solving a system of equations algebraically:EliminationSubstitution
3 EliminationThe key to solving a system by elimination is getting rid of one variable.Let’s review the Additive Inverse Property.What is the Additive Inverse of: 3x? -5y? 8p? q?-3x y -8p -qWhat happens if we add two additive inverses?We get zero. The terms cancel.We will try to eliminate one variable by adding, subtracting, ormultiplying the variable(s) until the two terms are additive inverses.We will then add the two equations, giving us one equation with one variable.Solve for that variable.Then insert the value into one of the original equations to find the other variable.
4 Elimination Solve the system: m + n = 6 m - n = 5 Notice that the n terms in both equations are additive inverses. So if we add the equations the n terms will cancel.So let’s add & solve: m + n = m - n = 52m + 0 = 112m = 11m = 11/2 or 5.5Insert the value of m to find n: n = 6n = .5The solution is (5.5, .5).
5 Elimination Solve the system: 3s - 2t = 10 4s + t = 6 We could multiply the second equation by 2 and thet terms would be inverses. ORWe could multiply the first equation by 4 and the second equationby -3 to make the s terms inverses.Let’s multiply the second equation by 2 to eliminate t. (It’s easier.)3s - 2t = s – 2t = 102(4s + t = 6) s + 2t = 12Add and solve: s + 0t = 2211s = 22s = 2Insert the value of s to find the value of t (2) - 2t = t = -2The solution is (2, -2).
6 Elimination Solve the system by elimination: 1. -4x + y = -12 4. 5m + 2n = -84m +3n = 2
7 Substitution To solve a system of equations by substitution… 1. Solve one equation for one of the variables.2. Substitute the value of the variable into the otherequation.Simplify and solve the equation.4. Substitute back into either equation to find the value of the other variable.
8 Substitution Solve the system: x - 2y = -5 y = x + 2 Notice: One equation is already solved for one variable.Substitute (x + 2) for y in the first equation.x - 2y = -5x - 2(x + 2) = -5We now have one equation with one variable. Simplify and solve.x - 2x – 4 = -5-x - 4 = -5-x = -1x = 1Substitute 1 for x in either equation to find y.y = x y = so y = 3The solution is (1, 3).
9 Substitution Let’s check the solution. The answer (1, 3) must check in both equations.x - 2y = y = x + 21 - 2(3) = = 1 + 2-5 = -5 3 = 3
10 Substitution Solve the system: 2p + 3q = 2 p - 3q = -17 Notice that neither equation is solved for a variable. Since p in thesecond equation does not have a coefficient, it will be easier to solve.p = 3q – 17Substitute the value of p into the first equation, and solve.2p + 3q = 22(3q – 17) + 3q = 26q – q = 29q – 34 = 29q = 36q = 4
11 Substitution 2p + 3q = 2 p – 3q = -17 Substitute the value of q into the second equation to find p.p = 3q – 17p = 3(4) – 17p = -5The solution is (-5, 4). (List p first since it comes first alphabetically.)Let’s check the solution:2p + 3q = 2 p – 3q = -172(-5) +3(4) = (4) = -17= = -172 = 2 = -17
12 Substitution Solve the systems by substitution: 1. x = 4 2x - 3y = -19