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Linear Equations 123
Solving Systems of Linear Equations by Addition and in Three Variables
| Question | Answer |
|---|---|
| Solve the System: {x+y=1 (1) {x-y=5 (2) | Add equations (1)+(2)to eliminate y. x=3 Solve for y. By plugging in x to either (1) or (2) y=-2 Solution is (3,-2) |
| Solve the System: {-2x+y=4 (1) {-x+3y=-8 (2) | Before adding equations must distribute -3 to (1) to eliminate y when adding (1)+(2), x=2.857 Solve for y. by plugging in x in either (1) or (2) y= 1.714 Solution is (2.857,-1.714) |
| Solve the System: {4x-2y=4 (1) {-8x+4y=-8 (2) | Before adding both equations must distribute 2 to (1) to eliminate y when adding (1)+(2), Solve for x. When solving for x everything ends up canceling out so the answer is 0=0 meaning infinite number of solutions. |
| Solve the System: {-x-(y/2)=7/2 (1) {(-x/2)+(y/4)=0 (2) | First must eliminate fractions, multiply (1) by 2, mulftiply (2) by 4. Add equations x=-4/7 Plug in x in either (1) or (2) to solve for y. y=-5.857 Solution is (-4/7,-5.857) |
| Solve the System: {2x-y+z=-8 (1) {x+2y-z=2 (2){2x+4y-2z=0 (3) | Add (1)+(2) eliminate z becomes (4) Then add (1)+(3) to eliminate z again becomes (5) Now (4)+(5) to eliminate y,solve for x x=-7/3 Replace x in(4)or(5) solve for, y=1 Finally replace x and y (1)(2)or(3) solve for z. Solution-(-7/3,1,7.3) |
| Solve the System: {2x-4y+8z=2 (1) {-x-3y+z=11 (2) {x-2y+4z=0 | Add (2)+(3) to eliminate x, becomes (4) Eliminate x again by (1)+(2) becomes (5) Using (4) and (5)solve for y and z. when you do this you end up getting the answer 0=2 This means the system is false and has no solution. |
| Solve the System: {4x+8y =1 (1) {-4x -4z=-1 (2) { y-4z=-3 (3) | Since there is no y in (2) add (1)+(3) to eliminate y. becomes equation 4. Next add (4)+(2) to solve for z. z=.893 plug that in to (3) and then solve for y. y=-.571 then use y to solve for x using equation (1). x=-.892 Solution is (.892,-.571,.893) |
| Solve the System: {x-y+z=-4 (1) {3x+2y-z=5 (2) {-2x+3y-z=15 (3) | Add (1)+(2) to eliminate z. becomes (4) Then add (1)+(3) to eliminate z again. becomes (5). add (4)and(5) to eliminate y,solve for x, x= -1. then use (4)or(5) to solve for y y=5. Then replace x,y in (1)(2)or(3) to solve for z. z=2 Solution is (-1,5,2 |
| Solve the system: {2x-4y+6z+-8 (1) {x-5y+5z=3 (2) {3x+y-3z=5 (3) | Add (1)+(2) to eliminate z. becomes(4) then add (1)+(3) to eliminate z. becomes (5) add(4)+(5) to eliminate y,solve for x. x=3 then solve for y using (4)or(5) y=2, use (1)(2)or(3) to solve for z using x,y. z=2 soultion (3,2,2) |
| Solve the system: {x+y+z=8 (1) {2x-y-z=10 (2) {x-2y-3z=22 (3) | Add (1)+(2) to eliminate z. becomes (4) then add (1)+(3) to eliminate z, becomes (5)Next add (4)+(5) to eliminate y & solve for x, x=6 replace x in (4) or (5) and solve for y. y=22. Then use (1)(2)(3) to solve for z using x,y z=-20 Solution: (6,22,-20) |
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