**In our previous post, we have seen how to solve linear equations using elimination method by addition or subtraction where the coefficients of one of the variable is either same or additive inverses.**

**Suppose in the given system of equations, the coefficient of either variables are not additive inverse or not the same then we have to use elimination method using multiplication to solve the system of linear equations.**

**In this method, we eliminate one variable by making the coefficients same by multiplying the equations by a constant.**

**Let us explain this method with some examples.**

**Example 1:**

**Solve the system of equations by elimination method.**

**3**

*x*+ 2*y*= 7**5**

*x*+ 4*y*= 11**Solution:**

**Step 1:**

**Given equations are 3**

*x*+ 2*y*= 7 ---- (1) and 5*x*+ 4*y*= 11 ---- (2).**Here, the coefficients of**

*x*and*y*are neither same nor additive inverses.**So, multiply equation (1) by 2 to make the coefficients of**

*y*same.**Equation (1) becomes 6**

*x*+ 4*y*= 14.**Step 2:**

**Now, subtract the equations.**

**6**

*x*+ 4*y*= 14**(-) 5**

*x*+ 4*y*= 11**----------------------**

*x*= 3**So,**

*x*= 3.**Step 3:**

**Substitute the value of**

*x*= 3 in either of the equations to find the value of*y*.**Let us substitute in equation (1).**

**3(3) + 2**

*y*= 7**9 + 2**

*y*= 7**Subtract 9 from both the sides and simplify.**

**9 + 2**

*y*- 9 = 7 - 9**2**

*y*= -2**Divide by 2 on both the sides to find the value of**

*y*.

*y*= -1**Step 4:**

**So, the solution for the given system of equations is (3, -1).**

**Note: In the above system of equations, we have multiplied one equation by a constant. In some cases, we need to multiply both the equations by a constant to make the coefficients same. Let us discuss this case in the next example.**

**Example 2:**

**Solve the system of equations using elimination method.**

**4**

*x*- 2*y*= -6**12**

*x*- 3*y*= -15**Solution:**

**Step 1:**

**Given equations are 4**

*x*- 2*y*= -6 ---- (1) and 12*x*- 3*y*= -15---- (2).**Here, the coefficients of**

*x*and*y*are not same or are not additive inverses.**So, we need to apply the elimination using multiplication method.**

**Step 2:**

**Multiply equation (1) by 3 on both the sides.**

**4**

*x*- 2*y*= -6 becomes 12*x*- 6*y*= -18.**And multiply equation (2) by -2 on both the sides.**

**12**

*x*- 3*y*= -15 becomes -24*x*- 6*y*= -30.**Now, observe that the coefficients of**

*y*in the two resultant equations, -6 and 6, are additive inverses.**Step 3:**

**Subtract the resultant two equations to find the value of**

*x*.**12**

*x*- 6*y*= -18**(-) -24**

*x*- 6*y*= -30 [-(24*x*- 6*y*= -30) = -24*x*+ 6*y*= 30]**-----------------------------**

**-12**

*x*= 12**Divide both the equations by -12 to isolate**

*x*.

*x*= -1**Step 4:**

**Now, substitute the value of**

*x*= -1 in either equation (1) or (2).**Let us substitute in equation (1).**

**4(-1) - 2**

*y*= -6**-4 - 2**

*y*= -6**Add 4 on both sides of the equation and simplify.**

**-4 - 2**

*y*+ 4 = -6 + 4**-2**

*y*= -2**Divide by -2 on both sides of the equation.**

*y*= 1**Step 5:**

**So, the solution of the given system of equations is (-1, 1).**

**Note: There are many other combinations of multipliers that can be used to multiply the equations to make the coefficients same**

**We can use two methods to solve a system of linear equations by elimination method using multiplication.**

**Method 1:**

**Step 1: Eliminate**

*y*.**Step 2: Find the value of**

*x*.**Step 3: Substitute the value of**

*x*and find the value of*y*.**Method 2:**

**Step 1: Eliminate**

*x*.**Step 2: Find the value of**

*y*.**Step 3: Substitute the value of**

*y*and find the value of*x*.**We have used method 1 in the above problem. We can also use method 2. Try to solve the above problem using the method 2.**

**The following table will guide us to choose the best method to solve.**

Methods | When to choose |

Graphing | When we are asked to estimate the solution. Graphing sometimes does not give the exact solution. |

Substitution | If the coefficient of one of the variables in either equation is 1 or -1. |

Elimination using addition | If one of the variables in the given two equations has the opposite coefficients (additive inverses). |

Elimination using subtraction | If one of the variables in the given two equations has the same coefficient. |

Elimination using multiplication | If none of the variables has the coefficients 1 or -1 and if the variables cannot be eliminated by using addition or subtraction. |

**Let us solve one more example on solving system of equations.**

**Example 3:**

**Choose the best method to solve the given system of equation and find the solution.**

**5**

*x*- 3*y*= 12**7**

*x*- 9*y*= 15**Solution:**

**Step 1:**

**Here, we need to find the exact solution of the given system. The algebraic method is the best way to find the exact solution.**

**None of the variables has coefficients 1 or -1. So, we cannot use substitution method.**

**The coefficients of**

*x*and*y*are neither same nor additive inverses. So, we cannot use the elimination using addition or subtraction. Hence, elimination using multiplication is the best method to solve the given system of equations.**Step 2:**

**Multiply the first equation by -3.**

**-15**

*x*+ 9*y*= -36**The second equation is 7**

*x*- 9*y*= 15.**Now, we can observe that the coefficients of**

*y*in the two equations are additive inverses.**Add the above two equations to eliminate**

*y*.**-15**

*x*+ 9*y*= -36**(+) 7**

*x*- 9*y*= 15**--------------------------**

**-8**

*x*= -21**Divide by -8 on both the sides to isolate**

*x*.

*x*=^{21}/_{8}**Step 3:**

**Substitute the value of**

*x*=^{21}/_{8}in the first equation 5*x*- 3*y*= 12 to find the value of*y*.**5(**

^{21}/_{8}) – 3y = 12

^{105}/_{8}– 3y = 12**Subtract**

^{105}/_{8}from both the sides of the equation and simplify.**(**

^{105}/_{8}) - 3*y*-(^{105}/_{8}) = 12 - (^{105}/_{8})**-3**

*y*= 12 - (^{105}/_{8})**-3y =**

^{(96 – 105)}/_{8}**-3**

*y*=^{-9}/_{8}**Divide by -3 on both the sides and simplify.**

*y*=^{3}/_{8}**Step 4:**

**So, the solution is (**

^{21}/_{8},^{3}/_{8}).**Practice Questions:**

**Solve the given system of equations using elimination method**

**1) 2x + 3y = 4; 3x + 2y = 1**

**2) 5x – y = -14; 2x + 4y = 12**

**3) -7x + 8y = -10; 4x – 3y = 12**

**4) 6x + 2y = -24; x + 7y = 16**

**5) -7x + 10y = 20; 8x – 5y = 35**

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