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solve the following simultaneous equation. 2x+2y+3z=210. 2x+3y+4z=270. 3x+4y+3z=300.​

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To solve the simultaneous equation: 2x+2y+3z=210, 2x+3y+4z=270, and 3x+4y+3z=300, we can use the elimination method. Here's how:

Step 1: Multiply the first equation by 2, and the second equation by -1 to eliminate x.4x + 4y + 6z = 420-2x - 3y - 4z = -270

Step 2: Add the two equations to eliminate x.2y + 2z = 150

Step 3: Multiply the first equation by -3, and the third equation by 2 to eliminate x.-6x - 6y - 9z = -6306x + 8y + 6z = 600

Step 4: Add the two equations to eliminate x.2y - 3z = -30

Step 5: Multiply the second equation by 2, and the fourth equation by 3 to eliminate y.4x + 6y + 8z = 540-6y + 9z = 90

Step 6: Add the two equations to eliminate y.4x + 17z = 630

Step 7: Substitute z = 2 into equation 2y + 2z = 150 to find y.2y + 4 = 150y = 73

Step 8: Substitute y = 73 and z = 2 into equation 4x + 17z = 630 to find x.4x + 34 = 630x = 149Therefore, the solution to the simultaneous equation 2x+2y+3z=210, 2x+3y+4z=270, and 3x+4y+3z=300 is x = 149, y = 73, and z = 2.

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