Answer:
The solution to the system of equations is:
x = 2, and y = -1
Explanation:
Given the pair of equations:
4x + 5y = 3 ..........................................................................(1)
2x + 3y = 1............................................................................(2)
To solve this by elimination:
Multiply equation (2) by 2, to eliminate x
Equation (2) becomes
4x + 6y = 2 .........................................................................(3)
Subtract equation (1) from (3)
4x - 4x + 6y - 5y = 2 - 3
y = -1 ....................................................................................(4)
Multiply equation (1) by 3 and equation (2) by 5 to eliminate y
Equation (1) becomes
12x + 15y = 9 .......................................................................(5)
Equation (2) becomes
10x + 15y = 5 ........................................................................(6)
Subtract equation (6) from (5)
12x - 10x + 15y - 15y = 9 - 5
2x = 4
Divide both sides by 2
x = 4/2 = 2 ............................................................................(7)
From equations (7) and (4)
x = 2, and y = -1
Answer:
y= -12
Step-by-step explanation:
4x+8 = -x+3 -> Find x by using all information
5x=-5 -> Simplify
x=-1 -> Solve
y=-12 -> Plug in to find y
We need to assign a value for x to check the possible values of y.
1st inequality: y < -0.75x
X = - 1 ; y < -0.75(-1) ; y < 0.75 possible coordinate (-1,0.75) LOCATED AT THE 2ND QUADRANT
X = 0 ; y < -0.75(0) ; y < 0 possible coordinate (0,0) ORIGIN
X = 1 ; y < -0.75(1) ; y < -0.75 possible coordinate (1,-0.75) LOCATED AT THE 4TH QUADRANT
2nd inequality: y < 3x -2
X = -1 ; y < 3(-1) – 2 ; y < -5 possible coordinate (-1,-5) LOCATED AT THE 4TH QUADRANT
X = 0 ; y < 3(0) – 2 ; y < -2 possible coordinate (0,-2) LOCATED AT THE 4TH QUADRANT
X = 1 ; y < 3(1) – 2 ; y <<span> 1 possible coordinate (1,1) LOCATED AT THE 1ST QUADRANT
The actual solution to the system lies on the 4TH QUADRANT.
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