solving systems of equations algebraically (solving systems of equations...
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SOLVING SYSTEMS OF EQUATIONS ALGEBRAICALLY (Solving Systems of Equations by Substitution and
Elimination) UNIT 03 LESSON 02
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OBJECTIVES
STUDENTS WILL BE ABLE TO:
Understand how liner system equations are solved by elimination and substitution method
KEY VOCABULARY:
• Solve linear equation and simultaneous equations.• Apply methods of substitution and elimination to
solve system of equations.
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3SOLVING SYSTEMS OF EQUATIONS ALGEBRAICALLY (BY SUBSTITUTION AND ELIMINATION)
INTRODUCTION
By system of equation we mean the system having
more than one equation.
A system of simultaneous equations is a group of
equations that must be all true at the same time.
By solution of equation we mean the values which
satisfy the given system of equations.
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4SOLVING SYSTEMS OF EQUATIONS ALGEBRAICALLY (BY SUBSTITUTION AND ELIMINATION)
INTRODUCTION
consider two linear equations in two variables, x and y, such as
5x - 3y = 43x + 3y = 1
Instead of one equation in one unknown, we have here two equations and two unknowns. In order to find a solution for this pair of equations, the unknown numbers x and y have to satisfy both equations.
Hence, we call this system or pair of equations or simultaneous equations. We now focus on various methods of solving simultaneous equations.
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5SOLVING SYSTEMS OF EQUATIONS ALGEBRAICALLY (BY SUBSTITUTION AND ELIMINATION)
METHODS TO SOLVE
Basically there are three methods to solve system of equations:
a. By substitution
b. By elimination
c. By comparison
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6SOLVING SYSTEMS OF EQUATIONS ALGEBRAICALLY (BY SUBSTITUTION AND ELIMINATION)
PROBLEM 01
Solve the system of equation by elimination
method.
7𝑥 + 2𝑦 = 47
5𝑥 − 4𝑦 = 1
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7SOLVING SYSTEMS OF EQUATIONS ALGEBRAICALLY (BY SUBSTITUTION AND ELIMINATION)
PROBLEM 01
7x+2y=47……………………(i)5x-4y=1……………………..(ii)
Multiply equation (i) by 2 then it becomes
14x+4y=94 ………………...(iii)
Now adding eq (ii) and (iii)5x-4y=114x+4y=94
19x = 95
𝒙 =𝟗𝟓
𝟏𝟗= 𝟓
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8SOLVING SYSTEMS OF EQUATIONS ALGEBRAICALLY (BY SUBSTITUTION AND ELIMINATION)
PROBLEM 01
By putting x=5 in eq ii
5 5 − 4𝑦 = 1
−4𝑦 = 1 − 25
4𝑦 = 24
𝑦 = 6
Solution Set= {(5,6)}
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9SOLVING SYSTEMS OF EQUATIONS ALGEBRAICALLY (BY SUBSTITUTION AND ELIMINATION)
PROBLEM 02
Solve the system of equation by elimination method.
𝑥 + 1
𝑦 + 1= 2
2𝑥 + 1
2𝑦 + 1=1
3
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10SOLVING SYSTEMS OF EQUATIONS ALGEBRAICALLY (BY SUBSTITUTION AND ELIMINATION)
PROBLEM 02
Simplify eq i and ii
𝑥 + 1 = 2(𝑦 + 1)
𝑥 = 2𝑦 + 2 − 1
𝑥 − 2𝑦 = 1………………..(iii)
3(2𝑥 + 1) = 2𝑦 + 1
6𝑥 + 3 = 2𝑦 + 1
6𝑥 − 2𝑦 = −2………..…..(iv)
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11SOLVING SYSTEMS OF EQUATIONS ALGEBRAICALLY (BY SUBSTITUTION AND ELIMINATION)
PROBLEM 02
Substring eq iii from iv we get
6x-2y=-2±x∓2y=±1
5x=-3x=-3/5
Substituting x=-3/5 in eq iii we get-3/5-2y=1-2y=1+3/5
y=2/5
Solution Set= {(-3/5,2/5)}
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12SOLVING SYSTEMS OF EQUATIONS ALGEBRAICALLY (BY SUBSTITUTION AND ELIMINATION)
PROBLEM 03
Solve the system of equation by substitution method
𝑥 + 𝑦 = 2
2𝑥 − 𝑦 = 1
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13SOLVING SYSTEMS OF EQUATIONS ALGEBRAICALLY (BY SUBSTITUTION AND ELIMINATION)
PROBLEM 03Solving equation (i)
𝑦 = 2 − 𝑥…………………(iii)
Substituting the value of y in equation (ii) we get,
2𝑥 − (2 − 𝑥) = 1
2𝑥 − 2 + 𝑥 = 1
3𝑥 = 3
𝒙 = 𝟏
Substituting the value of x in eq. iii.
𝑦 = 2 − 1
𝒚 = 𝟏
Solution set: {(x,y)}= {(1,1)}