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Solve each system of equation by using substitution 2j-3k=3 j+k=14

Sagot :

a386
[tex]2j-3k=3\\ j+k=14[/tex]

Get one variable by itself.  I chose j.  It is very simple to get j by itself in the second equation because all you have to do is move k to the right.

[tex]j+k=14\\ j+k-k=14-k\\j=14-k[/tex]

14-k is the new value for j.  Now you will only be working with the k variable.  Plug this value into the other equation for j.

[tex]2j-3k=3\\ 2(14-k)-3k=3[/tex]

Distribute the 2:
[tex]28-2k-3k=3[/tex]

Combine like terms and simplify:
[tex]28-5k=3\\ 28-5k-28=3-28\\ -5k=-25\\ \frac{-5k}{-5}=\frac{-25}{-5}\\k=5[/tex]

Now we have the value for k, which is 5.  We can plug this value into either equation to solve for j, but the second one is easier since we don't have to multiply 5 by a coefficient.

[tex]j+k=14\\j+5=14\\j+5-5=14-5\\j=9[/tex]

j=9
k=5
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