1 5 x 15 6 x 7 10 2 3 x 21 7 3 y 5 4 3 9 x 27 8 2 y 5 y 28 4 7 x 12 9 3 y 7 y

# 1 5 x 15 6 x 7 10 2 3 x 21 7 3 y 5 4 3 9 x 27 8 2 y 5

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1. 5x= 15 6. x+ 7 = 10 2. -3x= 21 7. 3y– 5 = 4 3. 9x= -27 8. 2y+ 5y= -28 4. -7x= -12 9. -3y+ 7y= 12 5. 23x= 8 10. 5x– 2x= -15QUESTIONS?a. How did you solve each equation?b. What mathematics concepts or principles did you apply to solve each equation? Explain how you applied these mathematics concepts and principles.c. Do you think there are other ways of solving each equation? Explain your answer.The solution of a system of linear equations can be determined algebraically or graphically. To find the solution graphically, graph both equations on a Cartesian coordinate plane then find the point of intersection of the graphs, if it exists. You may also use graphing calculator or computer software such as GeoGebra in determining the graphical solutions of systems of linear equations. GeoGebra is a dynamic mathematics software which helps you visualize and understand concepts in algebra, geometry, calculus, and statistics. The solution to a system of linear equations corresponds to the coordinates of the points of intersection of the graphs of the equations.Don’tForget!
302Don’tForget!Examples:Find the solutions of the following systems of linear equations graphically.a. 2x+ y =7-x+ y = 1b. 3x+ y =43xy = -5c. x– 2=-52x– 4y = -10 Answer (a): The graphs of 2x+ y= 7 and -x+ y= 1 intersect at (2, 3). Hence, the solution of the system 2x+ y =7-x+ y = 1is x= 2 and y = 3.Answer (b): The graphs of 3x+ y= 4 and 3x+ y= 10 are parallel. Hence, the system 3x+ y =43xy = -5has no solution.Answer (c): The graphs of x– 2y= -5 and 2x– 4y= -10 coincide. Hence, the system x– 2=-52x– 4y = -10 has infinite number of solutions.Teacher’s Note and Reminders
303A system of linear equations can be solved algebraically by substitutionorelimination methods.To solve a system of linear equations by substitution method, the following procedures could be followed:a. Solve for one variable in terms of the other variable in one of the equations.If one of the equations already gives the value of one variable, you may proceed to the next step.b. Substitute to the second equation the value of the variable found in the first step. Simplify then solve the resulting equation.c. Substitute the value obtained in (b) to any of the original equations to find the value of the other variable.d. Check the values of the variables obtained against the linear equations in the system.Example: Solve the system 2x+ y =5-x+ 2y = 5 by substitution method.Solution:Use 2x+ y= 5 to solve for yin terms of x.Subtract -2xfrom both sides of the equation.2x+ y– 2x= 5 – 2xy= 5 – 2xSubstitute 5 – 2xin the equation -x+ 2y= 5.-x+ 2(5 – 2x) = 5Simplify.-x+ 2(5) + 2(-2x) = 5 -x+ 10 – 4x= 5-5x= 5 – 10 -5x= -5Solve for xby dividing both sides of the equation by -5.

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