Solving Linear Systems by Substitution Lesson 3 2
Solving Linear Systems by Substitution Lesson 3. 2 Goal: Solve a system of linear equations in two variables by substitution
Warm-up 1. Evaluate -3 x – 5 y for x = -3 and y = 4. 2. Solve the system by graphing. x+y=2 2 x + y = 3 (1, 1) 3. Which system of linear equations has no solution? A. 7 x + y =10 B. 2 x – y =7 3 x – 2 y = -3 2 x – y = -7 C. 6 x – 2 y = 8 -3 x + y = -4 D. x + y =-5 6 x – 9 y = -15
Solving systems by substitution: Sometimes it is not practical or convenient to solve a system of equations by graphing. Therefore, another method to solve a system is by using substitution. Solving a Linear System by Substitution Step 1 Solve one equation for one of its variables. Step 2 Substitute the expression from Step 1 into the other equation and solve for the other variable. Step 3 Substitute the value from Step 2 into the revised equation from Step 1 and solve. Step 4 Check the solution in each of the original equations.
Example 1 – Solve the system This equation is already solved for y. Therefore, we can substitute 4 x for y into the second equation This is the value for x. Substitute it back into the other equation. Therefore, the solution for this system of equations is (-2, -8).
Example 2 – Solve the system It appears that it would be easier to solve for y in this equation. this for y in the 1 st Substitute equation. Solve for x. Therefore the solution is (-1,
Example 3 – Solve the system. Solve the 1 st equation for either x or y because you won’t have to deal with fractions. Substitute into the 2 nd equation The solution is (0, 3). Substitute this value into the other equation.
Example 4 In one week, a pizza restaurant sold 210 pizzas and made $2910. The price of a small pizza is $12 and the price of a large pizza is $15. How many small pizzas and large pizzas were sold that week? Let x = The number of small pizzas Let y = The number of large pizzas Therefore, they sold 80 small pizzas and 130 large pizzas.
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