Question

$\huge\red {QuEStiON}$



Solve using Cramer's Rule
3X +4y - 2z = 7
5X + y - 2z = 3
4X + 5y + 2z = 0​

Answer 1

Given:

Three equations-

3x+4y-2z=7

5x+y-2z=3

4x+5y+2z=0​

To Find:

Solution of given system of equation

Solution:

Let Δ be the determinant of coefficients of variables in the given system of equations

$\longrightarrow\sf{\triangle=\left|\begin{array}{ccc}3&4&-2\\5&1&-2\\4&5&2\end{array}\right|}$

$\longrightarrow\sf{\triangle=3(2+10)-4(10+8)-2(25-4)}$

$\longrightarrow\sf{\triangle=3(12)-4(18)-2(21)}$

$\longrightarrow\sf{\triangle=36-72-42}$

$\longrightarrow\sf{\triangle=-78}$

Now,

$\longrightarrow\sf{\triangle x=\left|\begin{array}{ccc}7&4&-2\\3&1&-2\\0&5&2\end{array}\right|}$

$\longrightarrow\sf{\triangle x=7(2+10)-4(6+0)-2(15-0)}$

$\longrightarrow\sf{\triangle x=7(12)-4(6)-2(15)}$

$\longrightarrow\sf{\triangle x=84-24-30}$

$\longrightarrow\sf{\triangle x=30}$

$\longrightarrow\sf{\triangle y=\left|\begin{array}{ccc}3&7&-2\\5&3&-2\\4&0&2\end{array}\right|}$

$\longrightarrow\sf{\triangle y=3(6+0)-7(10+8)-2(0-12)}$

$\longrightarrow\sf{\triangle y=3(6)-7(18)-2(-12)}$

$\longrightarrow\sf{\triangle y=18-126+24}$

$\longrightarrow\sf{\triangle y=-84}$

$\longrightarrow\sf{\triangle z=\left|\begin{array}{ccc}3&4&7\\5&1&3\\4&5&0\end{array}\right|}$

$\longrightarrow\sf{\triangle z=3(0-15)-4(0-12)+7(25-4)}$

$\longrightarrow\sf{\triangle z=3(-15)-4(-12)+7(21)}$

$\longrightarrow\sf{\triangle z=-45+48+147}$

$\longrightarrow\sf{\triangle z=150}$

Now, according to cramer's rule

$\longrightarrow\sf{x=\dfrac{\triangle x}{\triangle}}$

$\longrightarrow\sf{x=\dfrac{30}{-78}}$

$\longrightarrow\sf\green{x=\dfrac{-5}{13}}$

$\longrightarrow\sf{y=\dfrac{\triangle y}{\triangle}}$

$\longrightarrow\sf{y=\dfrac{-84}{-78}}$

$\longrightarrow\sf\green{y=\dfrac{14}{13}}$

$\longrightarrow\sf{z=\dfrac{\triangle z}{\triangle}}$

$\longrightarrow\sf{z=\dfrac{150}{-78}}$

$\longrightarrow\sf\green{z=\dfrac{-25}{13}}$

Answer 2

Answer:

To make the coefficients of the y's 4 and −4, we will multiply both sides of equation 2) by 4 :

1)   3x + 4y = 19 simultaneous equations 3x + 4y = 19

 

2)   2x − y = 9 simultaneous equations 8x − 4y = 36

 simultaneous equations

 11x     = 55

 

     x = 55

11

 

     x = 5

The 4 over the arrow in equation 2) signifies that both sides of that equation have been multiplied by 4.  Equation 1) has not been changed.

To solve for y, substitute  x = 5  in either one of the original equations.  In equation 1):

3· 5 + 4y = 19

 

4y = 19 − 15

 

4y = 4

 

y = 1

The solution is (5, 1).

The student should always verify the solution by replacing x and y with (5, 1) in the original equations.

Example 5.   Solve simultaneously:

1)   3x + 2y = −2

 

2)   2x + 5y = −5

Solution.   We must make one pair of coefficients negatives of one another.  In this example, we must decide which of the unknowns to eliminate, x or y.  In either case, we will make the new coefficients the Lowest Common Multiple (LCM) of the original coefficients -- but with opposite signs.  

Thus, if we eliminate x, then we will make the new coeffients 6 and −6.  (The LCM of 3 and 2 is 6.)  While if we eliminate y, we will make their new coefficients 10 and −10.  (The LCM of 2 and 5 is 10.)

Let us choose to eliminate x:

1)   3x + 2y = −2 simultaneous equations 6x + 4y = −4

 

2)   2x + 5y = −5 simultaneous equations −6x − 15y = 15

 ________________________________________________________________________

   − 11y = 11

     y = −1.

Equation 1) has been multiplied by 2.  Equation 2) has been multiplied by −3 -- because we want to make those coefficients  6 and −6, so that on adding, they will cancel.

To solve for x, we will substitute  y = −1  in the original equation 1):

3x + 2(−1) = −2

 

3x − 2 = −2

 

3x = 0

 

x = 0

The solution is (0, −1).

Problem 3.   Solve simultaneously.

1)   2x + 3y = 13

 

2)   5x − y = 7

To make the y's cancel, multiply equation 2) by 3:

1)   2x + 3y = 13 simultaneous equations 2x + 3y = 13

2)   5x − y = 7 simultaneous equations 15x − 3y = 21

 ________________________________________________________________________

 17x     = 34

     x = 2

To solve for y:

Substitute  x = 2  in one of the original equations.

In equation 1:

2· 2 + 3y = 13

 

4 + 3y = 13

 

3y = 9

 

y = 3

The solution is (2, 3).

Problem 4.   Solve simultaneously.

1)   x + 2y = −1

 

2)   2x − 3y = 5

To make the x's cancel, multiply equation 1) by −2:

1)   x + 2y = −1 simultaneous equations −2x − 4y = 2

 

2)   2x − 3y = 5 simultaneous equations 2x − 3y = 5

 ________________________________________________________________________

   − 7y = 7

 

     y = −1

To solve for x:

Substitute  y = −1  in one of the original equations.

In equation 1:

x + 2(−1) = −1

 

x − 2 = −1

 

x = −1 + 2

 

x = 1

The solution is (1, −1).

We could have eliminated y  by multiplying equation 1) by 3  and equation 2) by 2.

Problem 5.   Solve simultaneously:

1)   3x − 4y = 1

 

2)   2x + 3y = 12

To make the y's cancel:

Multiply equation 1) by 3  and equation 2) by 4:

1)   3x − 4y = 1 simultaneous equations 9x − 12y = 3

 

2)   2x + 3y = 12 simultaneous equations 8x + 12y = 48

 ________________________________________________________________________

 17x     = 51

 

     x = 51

17

 

     x = 3

To solve for y:

Substitute  x = 3  in one of the original equations.

In equation 2 (because the sign of y is already positive):

2· 3 + 3y = 12

 

6 + 3y = 12

 

3y = 6

 

y = 2

The solution is (3, 2).

Problem 6.   Solve simultaneously:

1)   3x + 2y = −4

 

2)   2x + 5y = 1

To make the x's cancel:

Multiply equation 1) by 2  and equation 2) by −3:

1)   3x + 2y = −4 simultaneous equations 6x + 4y = −8

 

2)   2x + 5y = 1 simultaneous equations −6x − 15y = −3

 ________________________________________________________________________

   − 11y = −11

 

     y = 1

To solve for x:

Substitute  y = 1  in one of the original equations.

In equation 1:

3x + 2· 1 = −4

 

3x + 2 = −4

 

3x = −4 − 2

 

3x = −6

 

x = −2

The solution is (−2, 1).

We could have eliminated y  by multiplying equation 1) by 5  and equation 2) by −2.

 

Step-by-step explanation: