QUESTION 4
The given system of linear equations are,
[tex]y = 5x - 7...(1)[/tex]
and
[tex] - 3x - 2y = - 12...(2)[/tex]
We put equation (1) into equation (2) to obtain,
[tex] - 3x - 2(5x - 7) = - 12[/tex]
We expand to get,
[tex] - 3x - 10x + 14 = - 12[/tex]
Group like terms to get,
[tex] - 3x - 10x = - 12 - 14[/tex]
Simplify to get,
[tex] - 13x = - 26[/tex]
Divide both sides by -13 to get,
[tex]x = 2[/tex]
Put the value of x into equation (1) to get,
[tex]y = 5(2) - 7[/tex]
[tex]y = 10 - 7 = 3[/tex]
The solution is
[tex]x = 2 \: \: and \: \: = 3[/tex]
QUESTION 5
The given system of linear equations are
[tex]-4x+y=6....(1)[/tex]
and
[tex]-5x-y=21...(2)[/tex]
We make y the subject in equation (1) to get,
[tex]y = 4x + 6....(3)[/tex]
Put equation (3) into equation (2) to get,
[tex] - 5x - (4x + 6) = 21[/tex]
We expand to get,
[tex] - 5x - 4x - 6 =21[/tex]
We group like terms to get,
[tex] - 5x - 4x = 21 + 6[/tex]
This implies that,
[tex] - 9x = 27[/tex]
We divide both sides by -9 to get,
[tex]x = - 3[/tex]
Put the value of x in to equation 3 to get,
[tex]y = 4( - 3) + 6[/tex]
[tex]y = - 12 + 6 = - 6[/tex]
Therefore the solution is
[tex]x = - 3 \: and \: y = - 6[/tex]
QUESTION 6
The given equations are,
[tex]-7x-2y=-13...(1)[/tex]
and
[tex]x-2y=-13...(2)[/tex]
Make x the subject in equation (2) to get,
[tex]x = 2y - 13...(3)[/tex]
Put equation (3) into equation (1) to get,
[tex] - 7(2y - 13) - 2y = - 13[/tex]
We expand to get,
[tex] - 14y + 91 - 2y = - 13[/tex]
We group like terms to obtain,
[tex] - 14y - 2y = - 13 - 91[/tex]
.Simplify to get,
[tex] - 16y = - 104[/tex]
Divide through by -16 to get,
[tex]y = \frac{13}{2} [/tex]
Put the value of y into equation (3) to get,
[tex]x = 2( \frac{13}{2} ) - 13[/tex]
[tex]x = 13 - 13 = 0[/tex]
Therefore
[tex]x = 0 \: and \: y = \frac{13}{2} [/tex]
