Question

How do you solve this set of linear equations: `\frac { 60} { x } + \frac { 60} { y } = \frac { 7} { 8};\frac { 40} { x } + \frac { 30} { y } = \frac { 1} { 2}`?

Answer 1

Isolate one of the variables to solve for substitution.

`(40(2y))/(2xy) + (30(2x))/(2xy) = (xy)/(2xy)`

We can now eliminate the denominators.

`80y + 60x = xy`

`80y + 60x - xy = 0`

`80y - xy = -60x`

`y(80 - x) = -60x`

`y = (-60x)/(80 - x)`

We can now substitute into the first equation.

`60/x + 60/((-60x)/(80 - x)) = 7/8`

`60/x + (60(80 - x))/(-60x) = 7/8`

`60/x - (80 - x)/(x) = 7/8`

`(60 - 80 + x)/x = 7/8`

`(-20 + x)/x = 7/8`

`8(-20 + x) = 7x`

`-160 + 8x = 7x`

`x = 160`

We can now substitute to solve for `y`.

`y = (-60x)/(80 - x)`

`y = (-60 xx 160)/(80 - 160)`

`y = -9600/-80`

`y = 120`

Hence, the solution set is `{160, 120}`.

Hopefully this helps!

Answer 2

`color(red)("Slightly different approach")`

`x=160`
`y=120`

Explanation:

`60/x+60/y=7/8...................Equation(1)`
`40/x+30/y=1/2.......................Equation(2)`
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
`color(blue)("Consider Equation(1)")`

`" "(60y+60x)/(xy)=7/8`

`" "7xy=8(60y+60x)...................Equation(1_a)`

`color(blue)("Consider Equation(2)")`

`" "(40y+30x)/(xy)=1/2`

`" "xy=2(40y+30x)............................Equation(2_a)`

`color(blue)("Substitute for "xy" in "Equation(1_a)" using "Equation(2_a))`

`" "14(40y+30x)=8(60y+60x)........Equation(1_a)`

`color(blue)("Divide both sides by 10")`

`" "14(4y+3x)=8(6y+6x)`

`color(blue)("Divide both sides by 2")`

`" "7(4y+3x)=4(6y+6x)`

`" "28y+21x=24y+24x`

`" "color(blue)(y=3/4x)`
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

Substitute for `y` in Equation(1)

`60/x+60/y=7/8" "->" "60/x+60/(3/4x) =7/8`

`1/x (60+(60xx4/3)) =7/8`

`color(blue)(" "x=160)`

`color(blue)("Thus "y=3/4xx160" "=120)`