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how would you solve 2x+3y=7x-4y=3?

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Guest Apr 11, 2017

Best Answer 

 #2
avatar+79894 
+2

2x+3y=7     (1)

x-4y=3    →  x  = 3 + 4y     (2)

 

Sub  (2)  into (1)   and we have

 

2 ( 3 + 4y) + 3y  = 7       simplify

 

6 + 8y + 3y   = 7

 

11y + 6   = 7        subtract  6 from both sides

 

11y  = 1             divide both sides by 11

 

y  = 1/11  

 

And using (2),    x  =  3 + 4(1/11)   =  3 + 4/11   =  37/11

 

So  (x, y)   =  ( 37/11, 1/11 )

 

 

cool cool cool 

CPhill  Apr 11, 2017
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3+0 Answers

 #1
avatar+7167 
+1

how would you solve 2x+3y=7x-4y=3?

 

\(2x+3y=3\\\underline{7x-4y=3}\\y= \frac{3-2x}{3}=\frac{3-7x}{7}\\21-14x=9-21x\\7x=-12\\\color{blue}x=-\frac{12}{7}\\\color{blue}y=\frac{3+\frac{24}{7}}{3}=\frac{15}{7}\)

 

\(-\frac{24}{7}+\frac{45}{7}=3\)

 

laugh  !

asinus  Apr 11, 2017
 #2
avatar+79894 
+2
Best Answer

2x+3y=7     (1)

x-4y=3    →  x  = 3 + 4y     (2)

 

Sub  (2)  into (1)   and we have

 

2 ( 3 + 4y) + 3y  = 7       simplify

 

6 + 8y + 3y   = 7

 

11y + 6   = 7        subtract  6 from both sides

 

11y  = 1             divide both sides by 11

 

y  = 1/11  

 

And using (2),    x  =  3 + 4(1/11)   =  3 + 4/11   =  37/11

 

So  (x, y)   =  ( 37/11, 1/11 )

 

 

cool cool cool 

CPhill  Apr 11, 2017
 #3
avatar+7167 
0

how would you solve 2x+3y=7x-4y=3?

 

\(2x+3y=3\\\underline{7x-4y=3}\)

 

\(\large y=\frac{3-2x}{3}\\\large y=\frac{7x-3}{4}\)                        y to the left

 

\(\large \frac{3-2x}{3}=\frac{7x-3}{4}\)                equilibrium

 

\(12-8x=21x-9\)           to common denominator 12

 

\(-29x=-21\)                     x to the lef,  rest to the right

 

\(\large x=\frac{21}{29}\)                              On both sides divided by -29

 

\(y=\frac{3-\frac{42}{29}}{3}\)                            x is used

 

\(\large y=\frac{15}{29}\)

 

\(2x+3y=3\)                        x and y are used

 

\(\large 2\times \frac{21}{29}+3\times \frac{15}{29}=3\)  the sample is correct

 

laugh  !

asinus  Apr 12, 2017

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