Question

What is the answer to the system of equations problem 3x=2y+10 9y=3x-7

Answer 1

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

$3x - 2y = 10\:(I)$
$3* \frac{76}{21} - 2* \frac{3}{7} = 10$
$\frac{228}{21} - \frac{6}{7} = 10$
$\frac{1596-126}{147} = 10$
$\frac{1470}{147} = 10$
$10 = 10\:(TRUE)$