For example, if your simultaneous equations are 1) x+2y=−4{\displaystyle x+2y=-4} and 2) 2x+5y=1{\displaystyle 2x+5y=1}, you will probably want to begin with the first equation, because the x{\displaystyle x} is already by itself.
For example, if you are beginning with x+2y=−4{\displaystyle x+2y=-4}, you could solve for x{\displaystyle x} by subtracting 2y from each side. x+2y=−4{\displaystyle x+2y=-4}x=−4−2y{\displaystyle x=-4-2y}
For example, if you found x=−4−2y{\displaystyle x=-4-2y} in the first equation, plug in −4−2y{\displaystyle -4-2y} for x{\displaystyle x} in the second equation:2x+5y=1{\displaystyle 2x+5y=1}2(−4−2y)+5y=1{\displaystyle 2(-4-2y)+5y=1}
For example, to solve for y{\displaystyle y} in the equation 2(−4−2y)+5=1{\displaystyle 2(-4-2y)+5=1}, first use the distributive property to multiply. 2(−4−2y)+5y=1{\displaystyle 2(-4-2y)+5y=1}−8−4y+5y=1{\displaystyle -8-4y+5y=1}−8+y=1{\displaystyle -8+y=1}y=9{\displaystyle y=9}
If you plug the y{\displaystyle y} value back into the second equation with the x{\displaystyle x} substitution, you will not be able to find the value of x{\displaystyle x}. [7] X Research source For example, if you found y=9{\displaystyle y=9}, plug in 9{\displaystyle 9} for y{\displaystyle y} in the first equation:x+2y=−4{\displaystyle x+2y=-4}x+2(9)=−4{\displaystyle x+2(9)=-4}
For example, to solve for x{\displaystyle x} in the equation x+2(9)=−4{\displaystyle x+2(9)=-4}, first multiply, and then subtract 18 from each side to find the value of x{\displaystyle x}. x+2(9)=−4{\displaystyle x+2(9)=-4}x+18=−4{\displaystyle x+18=-4}x=−22{\displaystyle x=-22}.
For example, if you found y=9{\displaystyle y=9} and x=−22{\displaystyle x=-22}, substitute these values into both equations. So, for the first equation:(−22)+2(9)=−4{\displaystyle (-22)+2(9)=-4}−22+18=−4{\displaystyle -22+18=-4}−4=−4{\displaystyle -4=-4} For the second equation:2(−22)+5(9)=1{\displaystyle 2(-22)+5(9)=1}−44+45=1{\displaystyle -44+45=1}1=1{\displaystyle 1=1}
