Problem to solve
Solve \[3[x-5y]+2[2x-1]=2[x-y]\], \[2[x-1]-3[x-y]=-3x\]
Solution
STEP 1
Solve the following system: {2 (2 x - 1) + 3 (-5 y + x) = 2 (-y + x) 2 (x - 1) - 3 (-y + x) = -3 x
STEP 2
Hint: | Expand the left hand side of the equation 2 (2 x - 1) + 3 (-5 y + x) = 2 (-y + x). 2 (2 x - 1) + 3 (-5 y + x) = (4 x - 2) + (-15 y + 3 x) = -15 y + 7 x - 2: {(-15 y + 7 x - 2) = 2 (-y + x) 2 (x - 1) - 3 (-y + x) = -3 x
STEP 3
Hint: | Expand the right hand side of the equation -15 y + 7 x - 2 = 2 (-y + x). 2 (-y + x) = +(-2 y + 2 x) = -2 y + 2 x: {-15 y + 7 x - 2 = (-2 y + 2 x) 2 (x - 1) - 3 (-y + x) = -3 x
STEP 4
Hint: | Expand the left hand side of the equation 2 (x - 1) - 3 (-y + x) = -3 x. 2 (x - 1) - 3 (-y + x) = (2 x - 2) + (3 y - 3 x) = 3 y - x - 2: {-15 y + 7 x - 2 = -2 y + 2 x (3 y - x - 2) = -3 x
STEP 5
Hint: | Choose an equation and a variable to solve for. In the second equation, look to solve for y: {-15 y + 7 x - 2 = -2 y + 2 x 3 y - x - 2 = -3 x
STEP 6
Hint: | Isolate terms with y to the left hand side. Add x + 2 to both sides: {-15 y + 7 x - 2 = -2 y + 2 x 3 y = -2 x + 2
STEP 7
Hint: | Solve for y. Divide both sides by 3: {-15 y + 7 x - 2 = -2 y + 2 x y = -(2 x)/3 + 2/3
STEP 8
Hint: | Perform a substitution. Substitute y = -(2 x)/3 + 2/3 into the first equation: {-2 - 15 (-(2 x)/3 + 2/3) + 7 x = 2 x - 2 (-(2 x)/3 + 2/3) y = -(2 x)/3 + 2/3
STEP 9
Hint: | Expand the left hand side of the equation -2 - 15 (-(2 x)/3 + 2/3) + 7 x = 2 x - 2 (-(2 x)/3 + 2/3). -2 - 15 (-(2 x)/3 + 2/3) + 7 x = (10 x - 10) + 7 x - 2 = 17 x - 12: {(17 x - 12) = 2 x - 2 (-(2 x)/3 + 2/3) y = -(2 x)/3 + 2/3
STEP 10
Hint: | Expand the right hand side of the equation 17 x - 12 = 2 x - 2 (-(2 x)/3 + 2/3). 2 x - 2 (-(2 x)/3 + 2/3) = ((4 x)/3 - 4/3) + 2 x = (10 x)/3 - 4/3: {17 x - 12 = ((10 x)/3 - 4/3) y = -(2 x)/3 + 2/3
STEP 11
Hint: | Choose an equation and a variable to solve for. In the first equation, look to solve for x: {17 x - 12 = (10 x)/3 - 4/3 y = -(2 x)/3 + 2/3
STEP 12
Hint: | Isolate x to the left hand side. Subtract (10 x)/3 - 12 from both sides: {(41 x)/3 = 32/3 y = -(2 x)/3 + 2/3
STEP 13
Hint: | Solve for x. Multiply both sides by 3/41: {x = 32/41 y = -(2 x)/3 + 2/3
STEP 14
Hint: | Perform a back substitution. Substitute x = 32/41 into the second equation: Answer: | | {x = 32/41 y = 6/41
More problems
Number
001Solve \(3x+2y=5\), \(-4x+5y=2\)
002Solve \(-4x+5y+3=x+4\), \(6x-7y=2y-1\)
003Solve \(-5x+2y=-3+2x\), \(4x-2y+3=7-4x\)
004Solve \(3y+5x+1-2y=2y-3\), \(3y-2+5x=2x-3y+2\)
005Solve \(3x+5y=2\), \(-5x+6y=5\)
006Solve \(-\frac{2}{3}x+5y=4\), \(7x-5y=4\)
007Solve \( 4x-5y=3\), \(-2x+6y=-2\)
008Solve \(\frac{6}{5}x+\frac{3}{2}y=5\), \(-\frac{3}{5}x+y=5y+2\)
009Solve \(\frac{3}{2}x-\frac{1}{2}y=6x-2\), \(3x+\frac{2}{3}y=-2y-1\)
010Solve \(-2x+6y-4x=5x-3y+3\), \(7y+4x=9-2y\)
011Solve \(\frac{2x+y}{3}-\frac{2(x+y)}{4}=2\), \(-\frac{2(x-y)}{5}-\frac{x+2y}{2}=-1\)
012Solve \(-\frac{2(x-y)}{5}+\frac{2x-y}{2}=-1\), \(4(2x+3)-\frac{3x-4y}{2}=5\)
013Solve \(\frac{x+2}{2}+\frac{3y-1}{4}=-1\), \(2x+y-2(3x-4y)=x+3\)
014Solve \(\frac{2x-1}{4}+\frac{x-2y}{3}=4+x\), \(\frac{x+y}{3}-(x-y)=-2y+1\)
015Solve \(3(x-5y)+2(2x-1)=2(x-y)\), \(2(x-1)-3(x-y)=-3x\)
016Solve \(x-y+3z=2\), \(2x-3y=-4\), \(x+y+z=-2\)
017Solve \(-4x+y-5z=2\), \(4x+2y+z=1\), \(2x-y+3z=4\)
018Solve \(-10x+3y+z=-2\), \(3x-2y+5z=4\), \(-6x+4y+2z=1\)
019Solve \(4x-3y=7\), \(2x-3y+z=7\), \(4x-y+z=-8\)
020Solve \(-x+3y+2z=5\), \(-3x-2y+z=-4\), \(x-y+z=3\)
021Solve \(\frac{x}{3}-\frac{y}{2}+z=2\), \(-x-\frac{y}{3}+2z=3\), \(2x-3y+\frac{z}{3}=-1\)
022Solve \(2x-\frac{y}{3}+2z=-1\), \(x+y+\frac{z}{2}=-1\), \(x-2y+z=2\)
023Solve \(-\frac{3x}{2}+3y+z=2\), \(3x-y+z=5\), \(\frac{x}{3}-y+z=2\)
024Solve \(\frac{2x}{3}-2z=-2\), \(2y-4z=2\), \(\frac{x}{3}-5y=2\)
025Solve \(\frac{x}{2}-3z=-2\), \(3y-\frac{z}{2}=2\), \(x+3y-z=5\)
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Edited by tododemate® - 2021