Question

solve the system of equations 2x+3y=17 3x-2y=6 by the method of cross multiplication​

Answer 1

Given :

• 2x + 3y = 17
• 3x - 2y = 6

To Find :

• The value of 'x' and 'y'

According to the question,

• 2x + 3y - 17 = 0
• 3x - 2y - 6 = 0

$: \implies\sf{ \dfrac{x}{(b_{1}c_{2} -b _{2}c_{1})} } = \dfrac{y}{(c_{1}a_{2} - c_{2}a_{1})} = \dfrac{1}{(a_{1}b_{2} - a_{2}b_{1})}$

$: \implies \sf{ \dfrac{x}{3 \times ( - 6) - ( - 2) \times ( - 17)} } = \dfrac{y}{( - 17) \times 3 - ( - 6) \times 2} = \dfrac{1}{2 \times ( - 2) - 3 \times 3}$

$: \implies\sf{ \dfrac{x}{ - 18 - 34} } = \dfrac{y}{ - 51 + 12} = \dfrac{1}{ - 4 - 9}$

$: \implies \sf{ \dfrac{x}{ - 52} = \dfrac{y}{ - 39} = \dfrac{1}{ - 13} }$

$: \implies \sf{ \dfrac{x}{ - 52} = \dfrac{1}{ - 13} } \longrightarrow{x = 4} \red \bigstar$

$: \implies \sf{ \dfrac{y}{ - 39} = \dfrac{1}{ - 13} } \longrightarrow y = 3 \: \blue \bigstar$

• Hence, the value of x = 4 and y = 3.

Answer 2

Answer:

Step-by-step explanation:

2x+3y=17                        ----[1]

3x-2y=6                           ---[2]

now write it as:-

2x+3y-17=0                      ----[3]

3x-2y-6=0                        ----[4]

now our values will be---

$a_{1} = 2\\b_{1}= 3\\c_{1} =-17$            

------------  

$a_{2}=3\\b_{2}=-2\\c_{2}= -6$

apply the values on the formula given below

$\frac{x}{c1b2-c2b1} = \frac{y}{b1a2-b2a1}=\frac{1}{a1c2-a2c1}$

after applying the values

$\frac{x}{(-17*-2)-(-6*3)} = \frac{y}{(3*3)-(-2*2)} = \frac{1}{(2*-6)-(3*-17)}$                       {note: first solve the                        

                                                                                                number  in the    

                                                                                                  brackets}  

now after getting the values

$\frac{x}{52}=\frac{y}{13} = \frac{1}{-39}$

now substitute the vaules of x and y with 1

$=>\frac{x}{52}= \frac{1}{-39} \\=>x= \frac{1}{-39} *52 =\frac{52}{-39} \\=>x= \frac{4}{3}$

------------------

$=>\frac{y}{13}= \frac{1}{-39} \\=>y=\frac{1}{39} *13=\frac{13}{-39} \\=>y=\frac{1}{3}$

ans==> x=4/3, y=1/3