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