How do you solve the following system: #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

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

# 10 y - 2x = 42 #

Use the commuative property to change the second equation

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

Now add the two equations

# 10y - 2x = 42#
# +-4y + 2x =6 # This gives

# 6y = 48 # divide both sides by 6

# 6y/6 = 48/6 # so

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

# 10xx 6 = 42 + 2x#

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

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

# 18 = 2x # divide each side by 2

# 18/2 = 2x/2 # so

# 9 = x #