SOLUTION: Could you please help me find the intercepts of the equation, and plot the graphs, showing the coordinates where the graphs intersect?
Here it is:
2x+y=8
5x-3y=15
Thanks s
Algebra ->
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-> SOLUTION: Could you please help me find the intercepts of the equation, and plot the graphs, showing the coordinates where the graphs intersect?
Here it is:
2x+y=8
5x-3y=15
Thanks s
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Question 23406: Could you please help me find the intercepts of the equation, and plot the graphs, showing the coordinates where the graphs intersect?
Here it is:
2x+y=8
5x-3y=15
Thanks so much for your help.
You can put this solution on YOUR website! SEE THE FOLLOWING EXAMPLE AND FOLLOW THE METHOD TO SOLVE YOUR PROBLEM.COME BACK IF NEEDED
INTERCEPTS ARE THE DISTANCE FROM ORIGIN TO THE POINT WHERE THE LINE CUTS THE X AXIS (CALLED X INTERCEPT);AND THE Y AXIS (CALLED Y INTERCEPT)ON THE X AXIS Y=0 .SO WE GET X INTERCEPT BY PUTTING Y=0 IN THE GIVEN EQN.FOR EX.IN 2X+Y=8,IF WE PUT Y=0 WE GET 2X=8 OR X=8/2=4.HENCE 4 IS THE X INTERCEPT.
ON Y AXIS X=0.SO WE GET Y INTERCEPT BY PUTTING X=0 IN THE GIVEN EQN.FOR EX.IN 2X+Y=8,IF WE PUT X=0 WE GET Y=8.HENCE 8 IS THE Y INTERCEPT
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Solve the system of linear equations by graphing: x+2y=5..
OR 2Y=5-X..OR Y=(5-X)/2
x-y=2...
OR Y=X-2
that is the first one
the second one is: 4x-y=6
3x+y=1
SOLVE BY GRAPHING MEANS DRAW THE GRAPHS AND FIND THE 2 GRAPHS OF 2 EQUATIONS CUT EACH OTHER .THAT IS THE SOLUTION.LET ME SHOW YOU THE FIRST PROBLEM. THEN YOU CAN DO SECOND PROBLEM THE SAME WAY
LET X=.........................0......1......2........3.........4......5....ETC.
Y=(5-X)/2..IN FIRST EQN........5/2....2......3/2......1.........1/2....0...ETC
Y=(X-2).....IN SECOND EQN.......-2....-1......0........1.........2......3...ETC
THE GRAPHS WILL LOOK LIKE THIS
WE FIND THAT THE TWO GRAPHS ARE CUTTING AT X=3 AND Y=1.HENCE THAT IS THE SOLUTION..YOU CAN CHECK BACK
EQN.I.....X+2Y=3+2*1=3+2=5..OK
EQN.II.....X-Y=3-1=2........OK.
