How do you solve the following system: $10 y = 42 + 2 x, 2 x - 4 y = 6$ $10y=42+2x , 2x-4y=6$ ?

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Answer 1 · 2017-07-13T12:52:08

Put both equations in the same form and add or subtract to eliminate x then solve for y

Change the first equation to put x and y on the same side

$10 y - 2 x = 42 + 2 x - 2 x$ $10y - 2x = 42 + 2x -2x$ gives

$10 y - 2 x = 42$ $10 y - 2x = 42$

Use the commuative property to change the second equation

$2 x - 4 y = 6 : - 4 y + 2 x = 6$ $2x - 4y = 6 : -4y + 2x = 6$

Now add the two equations

$10 y - 2 x = 42$ $10y - 2x = 42$
$\pm 4 y + 2 x = 6$ $+-4y + 2x =6$ This gives

$6 y = 48$ $6y = 48$ divide both sides by 6

$6 \frac{y}{6} = \frac{48}{6}$ $6y/6 = 48/6$ so

$y = 6$ $y = 6$ Put 6 in for y and solve for x

$10 \times 6 = 42 + 2 x$ $10xx 6 = 42 + 2x$

$60 = 42 + 2 x$ $60 = 42 + 2x$ subtract 42 from both sides

$60 - 42 = 42 - 42 + 2 x$ $60-42 = 42 -42 + 2x$ which gives

$18 = 2 x$ $18 = 2x$ divide each side by 2

$\frac{18}{2} = 2 \frac{x}{2}$ $18/2 = 2x/2$ so

$9 = x$ $9 = x$

How do you solve the following system: 10y=42+2x , 2x-4y=6 10y=42+2x,2x−4y=610y=42+2x , 2x-4y=6 ?