8th grade math

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 )

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

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

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

  !

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

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