Solve the following linear systems by ..... (elementary algebra)

Question

Solve the following linear systems by the method given alongside each. Verify your solutions also, in each case... (i). 2x-3y+z=1, x+y+z=2, 3x-4z-17=0 (by elimination method).. (ii). 2x-3y=1, 5-2y=z (by substitution method)... (iii). 2α+γ=3β, α+β=5+3γ, α-β=π (by Carmer's rule)... (iv). 3x-5=y, y=4 (geometrically)

Answer 1

(i) Elimination

A: 2x-3y+z=1; B: x+y+z=2; 3x-4z-17=0; C: 4z=3x-17.

D: A+3B: 5x+4z=7; 5x+(3x-17)=7; 8x=24, x=3; so 4z=9-17=-8, z=-2. y=2-x-z=2-3+2=1

CHECK: (x,y,z)=(3,1,-2); A: 6-3-2=1 OK; B: 3+1-2=2 OK; 3x-4z-17=9+8-17=0 OK.

(ii) Substitution

There appear to be three variable but only two equations, so a unique solution is not possible.

The solution will be expressed in terms of z. 2y=5-z; y=(5-z)/2; 2x-3y=1 becomes: 2x-3(5-z)/2=1; 4x-15+3z=2; 4x=17-3z, z=(17-3z)/4. (If z=3, x=2 and y=1, for example, or, if z=7, x=y=-1. Many solutions.)
(iii) Cramer's Rule

Let a = alpha, b=beta, c=gamma: 2a-3b+c=0; a+b-3c=5; a-b=(pi)

D=

| 2 -3 1 |

| 1 1 -3 | = 2(0-3)+3(3)+(-1-1)=-6+9-2=1

| 1 -1 0 |

Da=

| 0    -3  1 |

| 5     1 -3 | = 0+3(0+3(pi))+(-5-(pi))=8(pi)-5

| (pi) -1 0  |

Db=

| 2 0    1 |

| 1 5   -3 | = 2(0+3(pi))+((pi)-5)=7(pi)-5

| 1 (pi) 0 |

Dc=

| 2 -3 0    |

| 1  1 5    | = 2((pi)+5)+3((pi)-5)=5(pi)-5

| 1 -1 (pi) |

a=Da/D=8(pi)-5, b=Db/D=7(pi)-5, c=Dc/D=5(pi)-5

CHECK: 2a+c=16(pi)-10+5(pi)-5=21(pi)-15=3(7(pi)-5)=3b OK; a+b=15(pi)-10=5+15(pi)-15 OK; a-b=(pi) OK
(iv) Graphically (geometrically)

3x-5=y; y=4. y=4 is a horizontal line; y=3x-5 is a line with y intercept -5 and x intercept 5/3. Solution is (3,4), x=3, y=4.