7.4 for grade 10
Foundations & Pre-Calculus Mathematics 10 – 7.4
7.4 – USING A SUBSTITUTION STRATEGY TO SOLVE A LINEAR SYSTEM
Focus: Use the substitution of one variable to solve a linear system.
In the previous two classes, we solved linear system by graphing. A major shortcoming of this method is that at times we will only get an approximate solution, even if graphing technology is used.
We can use algebra to determine exact solutions to linear systems. One algebraic strategy is solving by substitution. By using substitution, we can transform a system of two linear equations into a single equation in one variable, which we can then solve as usual.
Consider the following linear system:
x − 4y + 4 = 0 i
4 x − y = 14 ii
In equation i, the variable x has coefficient 1. So, solve the equation for x.
Since the solution of the linear system is the point of intersection of the graphs of the two lines, the xcoordinate must satisfy both equations. We can substitute the expression for x in the other equation.
When we know the value of one variable, we substitute for that variable in one of the original equations then solve that equation for the other variable.
To verify the solution is correct, we substitute for both variables in the original equations.
Foundations & Pre-Calculus Mathematics 10 – 7.4
Example 1: Solving a Linear System by Substitution
2x − 4 y = 7
4x + y = 5
Example 2: Using a Linear System to Solve a Problem
a. Create a linear system to model this situation: Kimmo invested $1800, part at an annual interest rate
3.5% and the rest at an annual interest rate of 4.5%. After one year, the total interest was $73.
b. Solve this problem: How much money did Kimmo invest at each rate?
Foundations & Pre-Calculus Mathematics 10 – 7.4
Example 3: Solving a Linear System with Fractional Coefficients
1
4
x − y = −2
2
5
1
3 y= x−
4
8
Assignment: p. 425 #4 – 6, 8 – 10, 12 – 18 (even), 19, 23
7.4 – USING A SUBSTITUTION STRATEGY TO SOLVE A LINEAR SYSTEM
Focus: Use the substitution of one variable to solve a linear system.
In the previous two classes, we solved linear system by graphing. A major shortcoming of this method is that at times we will only get an approximate solution, even if graphing technology is used.
We can use algebra to determine exact solutions to linear systems. One algebraic strategy is solving by substitution. By using substitution, we can transform a system of two linear equations into a single equation in one variable, which we can then solve as usual.
Consider the following linear system:
x − 4y + 4 = 0 i
4 x − y = 14 ii
In equation i, the variable x has coefficient 1. So, solve the equation for x.
Since the solution of the linear system is the point of intersection of the graphs of the two lines, the xcoordinate must satisfy both equations. We can substitute the expression for x in the other equation.
When we know the value of one variable, we substitute for that variable in one of the original equations then solve that equation for the other variable.
To verify the solution is correct, we substitute for both variables in the original equations.
Foundations & Pre-Calculus Mathematics 10 – 7.4
Example 1: Solving a Linear System by Substitution
2x − 4 y = 7
4x + y = 5
Example 2: Using a Linear System to Solve a Problem
a. Create a linear system to model this situation: Kimmo invested $1800, part at an annual interest rate
3.5% and the rest at an annual interest rate of 4.5%. After one year, the total interest was $73.
b. Solve this problem: How much money did Kimmo invest at each rate?
Foundations & Pre-Calculus Mathematics 10 – 7.4
Example 3: Solving a Linear System with Fractional Coefficients
1
4
x − y = −2
2
5
1
3 y= x−
4
8
Assignment: p. 425 #4 – 6, 8 – 10, 12 – 18 (even), 19, 23