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solve by using the substitution method x+y=6 x+4y=3
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From the first equation, x=6-y. Substitute this value of x into the 2nd equation, you get: \[6-y+4y=3 \implies 6+3y=3 \implies 3y=-3 \implies y=-1\] So, we have y=-1. Now, substitute this value of y in equation (1) to get x: \[x-1=6 \implies x=7\] That's the solution of the system is is x=7 and y=-1
can you help me to sove some more ??please solve by the addition method 2x+y=11 x-y=11
Ok, Can you add equation (1) to equation (2)?
is 4+3 -6?
You should add the x's together, the y's together and the constants together. That would give you: 3x=22 ---> x=22/3 Substitute this value of x in equation (2) to find y.
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and i also need help to solve this one the last one 3x-2y=12 5x+2y=4
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