Solve the following pairs of linear equations by cross-multiplication method : 4x + 3y = 28; 9x - 5y = 63
Answers
Answer:
\textrm x=7,\textrm y=0x=7,y=0
Step-by-step explanation:
Given the pair of linear equation can be written as
4x+3y-28=0
9x-5y-63=0
these two equations are of the form ax+by+c=0
Here \textrm {a}_{1}=4, \textrm {b}_{1}=3, \textrm{c}_{1}=-28a
1
=4,b
1
=3,c
1
=−28
and \textrm{a}_{2}=9, \textrm{b}_{2}=-5,\textrm{c}_{2}=-63a
2
=9,b
2
=−5,c
2
=−63
We know from Cross multiplication,
\textrm {x}= \dfrac{\textrm b_{1}\textrm c_{2}-\textrm b_{2}\textrm c_{1}}{\textrm a_{1}\textrm b_{2}-\textrm a_{2}\textrm b_{1}}x=
a
1
b
2
−a
2
b
1
b
1
c
2
−b
2
c
1
and \textrm {y}= \dfrac{\textrm c_{1}\textrm a_{2}-\textrm c_{2}\textrm a_{1}}{\textrm a_{1}\textrm b_{2}-\textrm a_{2}\textrm b_{1}}y=
a
1
b
2
−a
2
b
1
c
1
a
2
−c
2
a
1
\textrm {x}= \dfrac{(3\times(-63)) -((-5)\times(-28))}{(4\times(-5))-(9\times3)}x=
(4×(−5))−(9×3)
(3×(−63))−((−5)×(−28))
\Rightarrow \textrm {x}= \dfrac{-189 -140}{-20-27}⇒x=
−20−27
−189−140
\Rightarrow \textrm {x}= \dfrac{-329}{-47}⇒x=
−47
−329
\Rightarrow \textrm x= 7⇒x=7
now \textrm {y}= \dfrac{((-28)\times9)-((-63)\times4)}{(4\times(-5))-(9\times3)}y=
(4×(−5))−(9×3)
((−28)×9)−((−63)×4)
\Rightarrow \textrm {y}= \dfrac{-252+252}{-20-27}⇒y=
−20−27
−252+252
\Rightarrow \textrm y=0⇒y=0
4x+3y/9x-5y=28/63
by cross multiplication
63(4x+3y)=28(9x-5y)
252x+189y=252x-140y
252x-252x= -189y-140y
x= -329