One way of solving a system of linear equations is by elimination, which is done by adding or subtracting the given equations to eliminate one of the variables.
This method is preferred when the coefficients of a variable are the same or can be made the same by eliminating one variable and reducing the system to a single equation.
Steps
Let us solve the system of linear equations:
3x + 2y = 12 …..(i)
5x – 2y = 8 …..(ii)
Step 1: Arranging the Equations in Standard Form
Both equations are written in the standard form:
a1x1 + a2x2 + … + anxn = b
Here,
- x1, x2, …, xn are the variables
- a1, a2, …, an are coefficients of the variables
- b is the constant term
In this case, both equations are already in standard form.
3x + 2y = 12 …..(i)
5x – 2y = 8 …..(ii)
Step 2: Identifying a Variable to Eliminate
If we consider the given equations, we get the y-terms that have the same coefficient with opposite signs:
Here, +2y is in equation (i), and -2y is in equation (ii).
Note: If the coefficients are unequal, we need to multiply one or both equations by suitable numbers to make the coefficients the same.
Step 3: Adding or Subtracting the Equations
- If the coefficients of a variable are the same, we subtract the equations.
- If the coefficients are opposites, we add the equations.
Here, the coefficients are opposite, which means we can eliminate y by adding the equations.
Now, adding both equations (i) and (ii), we get
(3x + 2y) + (5x – 2y) = 12 + 8
⇒ 3x + 2y + 5x – 2y = 20
Since +2y and -2y cancel out, we are left with:
⇒ 3x + 5x = 20
⇒ 8x = 20
Step 4: Solving for the Remaining Variable
Now, the remaining variable is x.
On dividing both sides of 8x = 20 by 8, we get
⇒ x = ${\dfrac{20}{8}}$
⇒ x = ${\dfrac{5}{2}}$
Step 5: Substituting x = ${\dfrac{5}{2}}$ Into One of the Original Equations
Now, substituting the value of x into equation (i), we get
3x + 2y = 12
⇒ ${3\left( \dfrac{5}{2}\right) +2y=12}$
⇒ ${\dfrac{15}{2}+2y=12}$
⇒ ${2y=12-\dfrac{15}{2}}$
⇒ ${2y=\dfrac{24-15}{2}}$
⇒ ${2y=\dfrac{9}{2}}$
⇒ ${y=\dfrac{9}{4}}$
Thus, the solution is: x = ${\dfrac{5}{2}}$ and y = ${\dfrac{9}{4}}$
Verifying
To verify the solutions, we will substitute both values into the original equations.
From equation (i),
3x + 2y = 12
⇒ ${3\left( \dfrac{5}{2}\right) +2\left( \dfrac{9}{4}\right) =12}$
⇒ ${\dfrac{15}{2}+\dfrac{9}{2}=12}$
⇒ ${\dfrac{15+9}{2}=12}$
⇒ ${\dfrac{24}{2}=12}$
⇒ 12 = 12, verified.
From equation (ii),
5x – 2y = 8
⇒ ${5\left( \dfrac{5}{2}\right) -2\left( \dfrac{9}{4}\right) =8}$
⇒ ${\dfrac{25}{2}-\dfrac{9}{2}=8}$
⇒ ${\dfrac{25-9}{2}=8}$
⇒ ${\dfrac{16}{2}=8}$
⇒ 8 = 8, verified.
Solved Examples
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Solve the system by elimination:
4x + 3y = 20
2x – 3y = 4
Solution:
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Given,
4x + 3y = 20 …..(i)
2x – 3y = 4 …..(ii)
Step 1: Arranging the Equations in Standard Form
Here, both equations are already in standard form.
Step 2: Identifying a Variable to Eliminate
Since the coefficients of y are already opposite (+3y and -3y), the two equations are added to eliminate y.
Step 3: Adding the Equations
(4x + 3y) + (2x – 3y) = 20 + 4
⇒ 4x + 3y + 2x – 3y = 24
⇒ 4x + 2x = 24
⇒ 6x = 24
Step 4: Solving for the Remaining Variable
On dividing both sides by 6,
⇒ x = 4
Step 5: Substituting x = 4 Into One of the Original Equations
Now, from (i),
4x + 3y = 20
⇒ 4(4) + 3y = 20
⇒ 16 + 3y = 20
⇒ 3y = 20 – 16
⇒ 3y = 4
⇒ y = ${\dfrac{4}{3}}$
Thus, the solution is: x = 4 and y = ${\dfrac{4}{3}}$
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Solve the system of equations using the elimination method:
7x – 3y = 31
5x – 3y = 9
Solution:
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Given,
7x – 3y = 31 …..(i)
5x – 3y = 9 …..(ii)
Step 1: Arranging the Equations in Standard Form
Here, both equations are already in standard form.
Step 2: Identifying a Variable to Eliminate
Since the coefficients of y are the same (-3y and -3y), the two equations are subtracted to eliminate y.
Step 3: Subtracting the Equations
(7x – 3y) – (5x – 3y) = 31 – 9
⇒ 7x – 3y – 5x + 3y = 22
⇒ 7x – 5x = 22
⇒ 2x = 22
Step 4: Solving for the Remaining Variable
On dividing both sides by 2,
⇒ x = 11
Step 5: Substituting x = 11 Into One of the Original Equations
Now, from (i),
7x – 3y = 31
⇒ 7(11) – 3y = 31
⇒ 77 – 3y = 31
⇒ -3y = 31 – 77
⇒ -3y = -46
⇒ 3y = 46
⇒ y = ${\dfrac{46}{3}}$
Thus, the solution is: x = 11 and y = ${\dfrac{46}{3}}$
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Solve:
3x + 2y = 8
5x – 4y = 10
Solution:
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Given,
3x + 2y = 8 …..(i)
5x – 4y = 10 …..(ii)
Step 1: Arranging the Equations in Standard Form
Here, both equations are already in standard form.
Step 2: Identifying a Variable to Eliminate
Since the coefficients of y are:
+2y in the first equation and -4y in the second equation.
This means to eliminate y, the coefficients of y are made equal by multiplying the equation (i) by 2 and equation (ii) by 1
Step 3: Multiplying the Equations to the Align Coefficients
2(3x + 2y) = 2(8) ⇒ 6x + 4y = 16
1(5x – 4y) = 1(10) ⇒ 5x – 4y = 10
Step 4: Adding the Equations
(6x + 4y) + (5x – 4y) = 16 + 10
⇒ 6x + 4y + 5x – 4y = 26
⇒ 6x + 5x = 26
⇒ 11x = 26
Step 5: Solving for the Remaining Variable
On dividing both sides by 11,
⇒ x = ${\dfrac{26}{11}}$
Step 6: Substituting x = ${\dfrac{26}{11}}$ Into One of the Original Equations
Now, from (i),
3x + 2y = 8
⇒ ${3\left( \dfrac{26}{11}\right) +2y=8}$
⇒ ${\dfrac{78}{11}+2y=8}$
⇒ ${2y=8-\dfrac{78}{11}}$
⇒ ${2y=\dfrac{88-78}{11}}$
⇒ ${y=\dfrac{10}{22}}$
⇒ ${y=\dfrac{5}{11}}$
Thus, the solution is: x = ${\dfrac{26}{11}}$ and y = ${\dfrac{5}{11}}$
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Solve by elimination:
3x + 4y = 14
5x – 2y = 16
Solution:
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Given,
3x + 4y = 14 …..(i)
5x – 2y = 16 …..(ii)
Step 1: Arranging the Equations in Standard Form
Here, both equations are already in standard form.
Step 2: Identifying a Variable to Eliminate
Since the coefficients of y are:
+4y in the first equation and -2y in the second equation.
This means to eliminate y, the coefficients of y are made equal by multiplying the equation (i) by 1 and equation (ii) by 2
Step 3: Multiplying the Equations to Align Coefficients
1(3x + 4y) = 1(14) ⇒ 3x + 4y = 14
2(5x – 2y) = 2(16) ⇒ 10x – 4y = 32
Step 4: Adding the Equations
(3x + 4y) + (10x – 4y) = 14 + 32
⇒ 3x + 4y + 10x – 4y = 46
⇒ 3x + 10x = 46
⇒ 13x = 46
Step 5: Solving for the Remaining Variable
On dividing both sides by 13,
⇒ x = ${\dfrac{46}{13}}$
Step 6: Substituting x = ${\dfrac{46}{13}}$ Into One of the Original Equations
Now, from (i),
${3\left( \dfrac{46}{13}\right) +4y=14}$
⇒ ${\dfrac{138}{13}+4y=14}$
⇒ ${4y=14-\dfrac{138}{13}}$
⇒ ${4y=\dfrac{182-138}{13}}$
⇒ ${4y=\dfrac{44}{13}}$
⇒ ${y=\dfrac{44}{52}}$
⇒ ${y=\dfrac{11}{13}}$
Thus, the solution is: x = ${\dfrac{46}{13}}$ and y = ${\dfrac{11}{13}}$
Problem: Solving for 3 VARIABLES
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Solve using the elimination method:
2x + y – z = 3
x – y + 3z = 7
3x + y + 2z = 4
Solution:
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Given,
2x + y – z = 3 …..(i)
x – y + 3z = 7 …..(ii)
3x + y + 2z = 4 …..(iii)
Step 1: Arranging the Equations in Standard Form
Here, both equations are already in standard form.
Step 2: Considering Two Equations to Eliminate One Variable
To eliminate y, the equations (i) and (iii) are subtracted:
(2x + y – z) – (3x + y + 2z) = 3 – 4
⇒ 2x + y – z – 3x – y – 2z = -1
⇒ 2x – 3x + y – y – z – 2z = -1
⇒ -x – 3z = -1
⇒ x + 3z = 1 …..(iv)
Similarly, eliminating y by adding the equations (i) and (ii),
(2x + y – z) + (x – y + 3z) = 3 + 7
⇒ 2x + y – z + x – y + 3z = 10
⇒ 2x + x + y – y – z + 3z = 10
⇒ 3x + 2z = 10 …..(v)
Step 3: Solving for One Variable
Solving equations (iv) and (v),
x + 3z = 1 …..(iv)
3x + 2z = 10 …..(v)
Now, multiplying equation (iv) by 3 and subtracting them,
3(x + 3z) – (3x + 2z) = 3(1) – 10
⇒ 3x + 9z – 3x – 2z = 3 – 10
⇒ 3x – 3x + 9z – 2z = -7
⇒ 7z = -7
⇒ z = -1
Step 4: Solving for x
From (iv),
x + 3z = 1
⇒ x = 1 – 3z
⇒ x = 1 – 3(-1)
⇒ x = 1 + 3
⇒ x = 4
Step 5: Solving for y
Substituting x = 4 and z = -1 into equation (i),
2x + y – z = 3
⇒ 2(4) + y – (-1) = 3
⇒ 8 + y + 1 = 3
⇒ y + 9 = 3
⇒ y = 3 – 9
⇒ y = -6
Thus, the solution is: x = 4, y = -6, and z = -1
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Solve by elimination:
2x + y – z = 3
x – 2y + 3z = 7
3x + y + 2z = 4
Solution:
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Given,
2x + y – z = 3 …..(i)
x – 2y + 3z = 7 …..(ii)
3x + y + 2z = 4 …..(iii)
Step 1: Arranging the Equations in Standard Form
Here, both equations are already in standard form.
Step 2: Considering Two Equations to Eliminate One Variable
To eliminate y, the equations (i) and (iii) are subtracted:
(3x + y + 2z) – (2x + y – z) = 4 – 3
⇒ 3x + y + 2z – 2x – y + z = 1
⇒ 3x – 2x + y – y + 2z + z = 1
⇒ x + 3z = 1 …..(iv)
Similarly, eliminating y by multiplying equation (i) by 2 and then adding the product by equation (ii),
2(2x + y – z) = 2(3)
⇒ 4x + 2y – 2z = 6 …..(v)
Adding equations (v) and (ii),
(4x + 2y – 2z) + (x – 2y + 3z) = 6 + 7
⇒ 4x + 2y – 2z + x – 2y + 3z = 13
⇒ 4x + x + 2y – 2y – 2z + 3z = 13
⇒ 5x + z = 13 …..(vi)
Step 3: Solving for One Variable
Now, solving equations (iv) and (vi):
x + 3z = 1 …..(iv)
5x + z = 13 …..(vi)
Multiplying (iv) by 5 and then subtracting from (vi),
(5x + z) – 5(x + 3z) = 13 – 5(1)
⇒ 5x + z – 5x – 15z = 13 – 5
⇒ 5x – 5x + z – 15z = 8
⇒ -14z = 8
⇒ z = ${-\dfrac{4}{7}}$
Step 4: Solving for x
From (iv),
x + 3z = 1
⇒ x = 1 – 3z
⇒ x = ${1-3\left( -\dfrac{4}{7}\right)}$
⇒ x = ${1+\dfrac{12}{7}}$
⇒ x = ${\dfrac{19}{7}}$
Step 5: Solving for y
Substituting x = ${\dfrac{19}{7}}$ and z = ${-\dfrac{4}{7}}$ into equation (i),
2x + y – z = 3
⇒ y = 3 – 2x + z
⇒ y = ${3-2\left( \dfrac{19}{7}\right) +\left( -\dfrac{4}{7}\right)}$
⇒ y = ${\dfrac{21-38-4}{7}}$
⇒ y = ${\dfrac{-21}{7}}$
⇒ y = -3
Thus, x = ${\dfrac{19}{7}}$, y = -3, and z = ${-\dfrac{4}{7}}$
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A science lab orders supplies for an experiment. Dr. Noah orders 3 beakers and 4 test tubes, spending a total of \$22. Dr. Kim purchases 5 beakers and 2 test tubes for \$18. Using the elimination method, determine the cost of one beaker and one test tube.
Solution:
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Let x be the cost of one beaker, and y be the cost of one test tube.
From the given problem,
Emma’s purchase is 3x + 4y = 22 …..(i)
Olivia’s purchase is 5x + 2y = 18 …..(ii)
Now, on multiplying equation (ii) by 2 to align the coefficients of y,
2(5x + 2y) = 2(18)
⇒ 10x + 4y = 36 …..(iii)
Subtracting equations (i) and (iii),
⇒ (3x + 4y) – (10x + 4y) = 22 – 36
⇒ 3x + 4y – 10x – 4y = -14
⇒ 3x – 10x = -14
⇒ -7x = -14
⇒ 7x = 14
⇒ x = 2
Now, substituting x = 2 into the equation (i),
3x + 4y = 22
⇒ 3(2) + 4y = 22
⇒ 6 + 4y = 22
⇒ 4y = 22 – 6
⇒ 4y = 16
⇒ y = 4
Thus, one beaker costs \$2, and one test tube costs \$4.
