How to solve this kind of equation? y = 3x y = -7x+2
its not an equation its a system of two equations
You can substitute the first equation into the second equation, \[3x=-7x+2\] solve for x then plug back into the first equation to solve for y :P
@ganeshie8 That doesn't help anyone.
loll
how are you supposed to reorganize them to do the math
If you're given more than one equation, then it's a system of equations problem
well no, these are two seperate equations, they're just examples
y = 3x is not in any way connected to y = -7x2
er, -7x+2
ohk... got you, its not a system. do you know how the graph of any of the given equations looks like ?
its not a question asking for a graph, its asking for a pair of numbers that work for the equation
ie. 2, 9, 4, 4, 5,3, etc...
Okay, you may do this : 1) plug any number for x in the equation 2) solve y you get an ordered pair that is a solution of the equation
what is the specific formula for finding out what the pair of numbers is
there are no "the pair of numbers" the solution of a linear equation, like the ones you have, is a set of infinitely many ordered pairs. it's hard to give you a visual picture w/o refering to the graph
there isn't a graph. this is the exact question. "Which ordered pair is a solution of the equation y = 3x? -2, -9 -8, -18 -8, -3 -10, -30"
y = 3x any pair of numbers that satisfy above equation is "a" solution
Look at the options one by one. To test if `-2, -9` is a solution, simply plugin x = -2 and y = -9 in the equation and see if it satisfies : y = 3x -9 = 3(-2) -9 = -6 which is clearly false, so -2, -9 is not a solution check next option
okay nevermind, i got it. i didn't remember that the equations backwards and the first number is X and the second is Y
Yes, it's always (x,y) even when you read a graph look at the horizontal than vertical
It is a widely followed convention for an ordered pair, the first number is always the \(x\) coordinate : \((x, y)\)
When possible, please try to post the full question. It helps fellow helpers in understanding the actual question quick and responding accordingly...
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