Solving:
[tex] \left \{ {{3x = 2y + 10} \atop {9y = 3x-7}} \right. [/tex]
Organize by putting the unknowns left and right the numbers without letter, changing the signal when changing sides.
[tex] \left \{ {{3x-2y=10\:(I)} \atop {-3x+9y=-7\:(II)}} \right. [/tex]
Eliminate opposites (+ 3x and -3x)
[tex]\left \{ {{\diagup\!\!\!\!3x-2y=10} \atop {-\diagup\!\!\!\!3x+9y=-7}} \right. [/tex]
[tex]\left \{ {{-2y=10} \atop {9y=-7}} \right. [/tex]
----------------------------
[tex]7y = 3[/tex]
[tex]\boxed{y = \frac{3}{7} }[/tex]
Now substitute the found value "y" in the first equation:
[tex]3x-2y=10\:(I)[/tex]
[tex]3x-2* \frac{3}{7} =10[/tex]
[tex]3x- \frac{6}{7} = 10[/tex]
[tex] \frac{21x}{\diagup\!\!\!\!7} - \frac{6}{\diagup\!\!\!\!7} = \frac{70}{\diagup\!\!\!\!7} [/tex]
[tex]21x - 6 = 70[/tex]
[tex]21x = 70 + 6[/tex]
[tex]21x = 76[/tex]
[tex]\boxed{x = \frac{76}{21} }[/tex]
Answer:
[tex](x,y) = ( \frac{76}{21} , \frac{3}{7} )\end{array}}\qquad\quad\checkmark[/tex]
Taking the truth proof
[tex]3x - 2y = 10\:(I)[/tex]
[tex]3* \frac{76}{21} - 2* \frac{3}{7} = 10[/tex]
[tex] \frac{228}{21} - \frac{6}{7} = 10[/tex]
[tex] \frac{1596-126}{147} = 10[/tex]
[tex] \frac{1470}{147} = 10[/tex]
[tex]10 = 10\:(TRUE)[/tex]
