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how to solve linear equations by substitution?

5x+2y=9
x+y=-3


Sagot :

[tex]\begin{cases} 5x+2y=9 \\ x+y=-3\end{cases} \\ \\\begin{cases} 5x+2*(-3-x)=9 \\ y=-3 -x\end{cases} \\ \\\begin{cases} 5x-6-2x =9 \\ y=-3 -x\end{cases}\\ \\\begin{cases} 3x =9+6 \\ y=-3 -x\end{cases}[/tex]

[tex]\begin{cases} 3x =15 \ \ /:3 \\ y=-3 -x\end{cases} \\ \\\begin{cases} x =5 \\ y=-3 -5\end{cases} \\ \\\begin{cases} x =5 \\ y=-8\end{cases}[/tex]


5x+2y=9 x+y=3

change x+y=3 into x=3-y

plug that equation into 5x+2y=9 where x is.

5(3-y)+2y=9

then distribute.

15-5y+2y=9

15-7y=9

minus 15.

-7y=-6

divide both sides by -7.

y is equal to -6/7.

then go back to the first equation and plug in for y.

and then you will get x.

then you will have (y,x)