Answer:
[tex]x = \dfrac{110}{31}; \qquad y = \dfrac{72 }{31}[/tex]
Step-by-step explanation:
I am guessing that your two equations are
(1) ⅕x + ⅛y = 1
(2) ½x - ⅓ y = 1
To get rid of fractions, I would multiply each equation by the least common multiple of its denominators.
[tex]\begin{array}{rcrl}(3) \qquad 8x + 5y & = & 40 & \text{Multiplied (1) by 40}\\(4) \qquad 3x - 2y & = & 6 & \text{Multiplied (2) by 6}\\\end{array}[/tex]
We can solve this system of equations by the method of elimination.
[tex]\begin{array}{rcrl}(5) \qquad \, \, 16x + 10y & = & 80 & \text{Multiplied (3) by 2}\\(6) \qquad \, \: 15x - 10 y & = & 30 & \text{Multiplied (4) by 5}\\31x & = & 110 & \text{Added (5) and (6)}\\\\(7)\qquad\qquad \qquad x & = & \dfrac{110 }{31} & \text{Divided each side by 31}\\\end{array}[/tex]
[tex]\begin{array}{rcrl}3 \left (\dfrac{110}{31} \right) - 2y & = & 6 & \text{Substituted (7) into (4)}\\\\ (5) \qquad16x + 10y & = & 80 & \text{Multiplied (3) by 2}\\\\(6)\qquad 15x - 10 y & = & 30 & \text{Multiplied (4) by 5}\\\\31x & = & 110 & \text{Added (5) and (6)}\\\\(7)\qquad \qquad \qquad x & = & \dfrac{110 }{31} & \text{Divided each side by 31}\\\\3 \left(\dfrac{110}{31} \right ) - 2y & = & 6 & \text{Substituted (7) into (4)}\\\\\end{array}\\\\[/tex]
[tex]\begin{array}{rcll}\dfrac{330}{31} - 2y & = & 6 &\\\\-2y & = & 6 - \dfrac{330}{31} &\\\\y & = & \dfrac{165}{31} -3 & \text{Divided each side by -2}\\\\ & = & \dfrac{165 - 93}{31} &\\\\ & = & \dfrac{72}{31} &\\\\\end{array}\\\\\therefore x = \dfrac{110}{31}; \qquad y = \dfrac{72 }{31}[/tex]
The diagram below shows the graphs of your two functions intersecting at (3.548, 2.323). These are the decimal equivalents of your fractional coordinates.
