6. Calculate 3 z-18 (mod 4 ) 7. Solve the Simultaneous equations by elimination methos [ 3 y-4 x=8 ( and ) 2 y-x=9 ] 8. find the tion set of the following simultateous equation. (a) (2)/(3) x-(2)/(4) y=0 and (1)/(3) x+(1)/(2) y=7 S. (1)/(4) k-(1)/(9) l=7 and (1)/(4) k-(1)/(9) l=1 9. The sum of two numbers is 8 and their product is -33 find the two numbers 10 A bill of 135 naira was paid wing 5 naira note and 10 naira note. If 16 notes were used atogethes, find howmany of each were used.

6. To calculate \( 3z - 18 \) (mod 4), we need to find the remainder when \( 3z - 18 \) is divided by 4.<br /><br />First, let's simplify the expression:<br />\[ 3z - 18 \equiv 3(z - 6) \pmod{4} \]<br /><br />Since \( z - 6 \) is an integer, we can write it as \( 4k + r \) where \( k \) is an integer and \( r \) is the remainder when \( z - 6 \) is divided by 4.<br /><br />So, \( z - 6 \equiv 4k + r \pmod{4} \).<br /><br />Now, substitute this back into the expression:<br />\[ 3(z - 6) \equiv 3(4k + r) \equiv 12k + 3r \pmod{4} \]<br /><br />Since \( 12k \) is divisible by 4, we have:<br />\[ 12k \equiv 0 \pmod{4} \]<br /><br />So, the expression simplifies to:<br />\[ 3(z - 6) \equiv 3r \pmod{4} \]<br /><br />Since \( r \) can be any integer from 0 to 3, the possible values for \( 3r \) are 0, 3, 6, and 9.<br /><br />Therefore, \( 3z - 18 \equiv 0, 3, 6, \text{ or } 9 \pmod{4} \).<br /><br />7. To solve the simultaneous equations by elimination method, we can multiply the second equation by 4 to make the coefficients of \( x \) in both equations equal:<br /><br />\[ 3y - 4x = 8 \]<br />\[ 8y - 4x = 36 \]<br /><br />Now, subtract the first equation from the second equation to eliminate \( x \):<br />\[ (8y - 4x) - (3y - 4x) = 36 - 8 \]<br />\[ 5y = 28 \]<br />\[ y = \frac{28}{5} \]<br /><br />Substitute \( y = \frac{28}{5} \) into the first equation to solve for \( x \):<br />\[ 3(\frac{28}{5}) - 4x = 8 \]<br />\[ \frac{84}{5} - 4x = 8 \]<br />\[ -4x = 8 - \frac{84}{5} \]<br />\[ -4x = \frac{40}{5} - \frac{84}{5} \]<br />\[ -4x = -\frac{44}{5} \]<br />\[ x = \frac{44}{20} \]<br />\[ x = \frac{11}{5} \]<br /><br />So, the solution to the system of equations is \( x = \frac{11}{5} \) and \( y = \frac{28}{5} \).<br /><br />8. (a) To solve the system of equations \( \frac{2}{3}x - \frac{2}{4}y = 0 \) and \( \frac{1}{3}x + \frac{1}{2}y = 7 \), we can multiply the first equation by 6 and the second equation by 6 to eliminate the fractions:<br /><br />\[ 4x - 3y = 0 \]<br />\[ 2x + 3y = 42 \]<br /><br />Now, add the two equations to eliminate \( y \):<br />\[ (4x - 3y) + (2x + 3y) = 0 + 42 \]<br />\[ 6x = 42 \]<br />\[ x = 7 \]<br /><br />Substitute \( x = 7 \) into the first equation to solve for \( y \):<br />\[ \frac{2}{3}(7) - \frac{2}{4}y = 0 \]<br />\[ \frac{14}{3} - \frac{1}{2}y = 0 \]<br />\[ \frac{1}{2}y = \frac{14}{3} \]<br />\[ y = \frac{28}{3} \]<br /><br />So, the solution to the system of equations is \( x = 7 \) and \( y = \frac{28}{3} \).<br /><br />(b) To solve the system of equations \( \frac{1}{4}k - \frac{1}{9}l = 7 \) and \( \frac{1}{4}k - \frac{1}{9}l = 1 \), we can multiply both equations by 36 to eliminate the fractions:<br /><br />\[ 9k - 4l = 252 \]<br />\[ 9k - 4l = 36 \]<br /><br />Now, subtract the second equation from the first equation to eliminate \( l \):<br />\[ (