\left\{ \begin{array} { l } { ( 5 x + 1 ) - 4 ( 5 \times 30 ) + 40 J = 16 } \\ { ( 3 x + 3 y ) - [ ( 6 \times 30 ) + ( 3 \times 40 ) ] = 120 } \end{array} \right.
Solve for x, y
x=123-8J
y=8J+17
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5x+40J-599=16,3x+3y-300=120
To solve a pair of equations using substitution, first solve one of the equations for one of the variables. Then substitute the result for that variable in the other equation.
5x+40J-599=16
Pick one of the two equations which is more simple to solve for x by isolating x on the left hand side of the equal sign.
5x=615-40J
Subtract -599+40J from both sides of the equation.
x=123-8J
Divide both sides by 5.
3\left(123-8J\right)+3y-300=120
Substitute 123-8J for x in the other equation, 3x+3y-300=120.
369-24J+3y-300=120
Multiply 3 times 123-8J.
3y+69-24J=120
Add 369-24J to -300.
3y=24J+51
Subtract 69-24J from both sides of the equation.
y=8J+17
Divide both sides by 3.
x=123-8J,y=8J+17
The system is now solved.
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