Question

How do you solve the system of equations `\frac { 3} { 4} x + \frac { 1} { 3} y = 1` and `\frac { 2} { 5} x - \frac { 1} { 4} y = \frac { 31} { 10}`?

Answer 1

`x=4 and y=-6`

Explanation:

`\frac { 3} { 4} x + \frac { 1} { 3} y = 1` `->`let this be equation (`1`),

and

`\frac {2} {5} x-\frac {1} {4}y = \frac {31} {10}` `->`Let this be equation (`2`)

Make coefficients of any one variable same.

As the sign of variable `y` is opposite in both the equations, if we make the coefficients of `y` equal and add the equations, `y` will be eliminated and we can get value of `x`.

So we multiply (`1`) by `1/4` and (`2`) by `1/3` and add the resulting equations:

(`1`)`xx 1/4 =>`

`\frac {3} {4}x xx(1/4) + \frac {1} {3}y xx(1/4) = 1 xx (1/4)`

`=> (3x)/16 + y/12 = 1/4` `->` (`1'`)

(2)`xx 1/3 =>`

`\frac{2} {5}x xx(1/3)-\frac {1}{4}y xx(1/3) = \frac{31}{10}xx(1/3)`

`=> (2x)/15 -y/12=31/30` `->` (`2'`)

Now adding (1') and (2') `=>`

`=> (3x)/16 +cancel( y/12) +(2x)/15 -cancel(y/12) = 1/4+31/30`

`=> (3x)/16(15/15) +(2x)/15 (16/16)= 1/4 xx (15/15)+31/30(2/2)`

`=> (45x + 32x)/240 = (15+62)/60`

`=> (77x)/240 = 77/60`

`=> x= 77/60 xx 240/77`

`therefore x= 4`

Now we can substitute this value of `x` in any one initial equations and find value of `y`:

(1) `=> \frac {3} {cancel4} xx cancel4 + \frac { 1} { 3} y = 1`

`=> 3 +1/3 y = 1`

`=> 1/3 y = 1-3 =-2`

`=> y= -2xx3= -6`

`therefore y= -6`

Answer: `x=4 and y=-6`

Answer 2

`x=4 and y=-6`

Explanation:

The first thing to do is to get rid of the fractions so you can work with whole numbers.

Multiply each equation by the LCM of the denominators so that you can cancel them.

`(color(red)(cancel12^3xx)3x)/cancel4+(color(red)(cancel12^4xx)1y)/cancel3=color(red)(12xx)1" "rarr" "9x+4y =12`

`(color(blue)(cancel20^4xx)2x)/cancel5(color(blue)(cancel20^5xx)1y)/cancel4= (color(blue)(cancel20^2xx)31)/cancel10" "rarr" "8x-5y =62`

This gives two much simpler equations to solve. Make the `y` terms into additive inverses so that they can be eliminated by adding.

`color(white)(xxxxxxxxxx)9x+4y = 12color(white)(xxxxxxxxxxxxx) A`
`color(white)(xxxxxxxxxx)8x-5y = 62color(white)(xxxxxxxxxxxxx) B`

`A xx5color(white)(xxxxx)45x+20y = 60color(white)(xxxxxxxxxxxx) C`
`Bxx4color(white)(xxxxx)32x-20y = 248color(white)(xxxxxxxxx) D`

`C+Dcolor(white)(xxxxxx)77xcolor(white)(xx)=308`

`color(white)(xxxxxxxxxxxx)xcolor(white)(xx)=308/77 = 4`

Substitute into `A` to find `y`

`color(white)(xxxxxxxxxx)9(4)+4y = 12`

`color(white)(xxxxxxxxxxx)36+4y = 12`

`color(white)(xxxxxxxxxxxxxxx)4y = 12-36 =-24`

`color(white)(xxxxxxxxxxxxxxxx)y = -6`

Check:

`2/5x-1/4y=31/10`

`2/5(4)-1/4(-6)=31/10`

`8/5+6/4 =31/10`

`31/10=31/10`