Question 149211
The basics:
You have 2 equations and 2 unknowns x, and y
In that situation, you can ALWAYS solve for the unknowns
Even if you had 30 equations and 30 unknowns, it would be
more difficult, but you could still solve for each of
your unknowns
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In this problem you can use the simple laws of algebra
to solve for x and y, mainly:
Whatever you do to one side of the equation, you MUST
do the same thing to the other side in order for the
equation to still be true
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{{{-x + y = 4}}}
Add {{{x}}} to both sides
{{{-x + x + y = 4 + x}}}
On the left side, the {{{x}}}'s cancel
{{{y = 4 + x}}}
That's what I want: one of the unknowns all by itself
Now I can replace {{{y}}} in the other equation with
{{{4 + x}}} because it equals {{{y}}}
{{{2x - 5*(4 + x) = -14}}}
Now I carry out the multiplication
{{{2x - 20 - 5x = -14}}}
{{{-3x - 20 = -14}}}
Multiply both sides by {{{-1}}}
{{{3x + 20 = 14}}}
Subtract {{{20}}} from both sides
{{{3x = -6}}}
{{{x = -2}}}
Now substitute this value for {{{x}}} back into either
of the equations
{{{-x + y = 4}}}
{{{-(-2) + y = 4}}}
{{{2 + y = 4}}}
Subtract {{{2}}} from both sides
{{{y = 2}}}
The solutions are x = -2 and y = 2
Now check the answers:
{{{-x + y = 4}}}
{{{-(-2) + 2 = 4}}}
{{{2 + 2 = 4}}}
{{{4 = 4}}}
and
{{{2x - 5y = -14}}}
{{{2*(-2) - 5*2 = -14}}}
{{{-4 - 10 = -14}}}
{{{-14 = -14}}}
OK