{"id":2895,"date":"2023-10-03T16:00:50","date_gmt":"2023-10-03T16:00:50","guid":{"rendered":"http:\/\/localhost\/ecole9ja\/?p=2895"},"modified":"2023-10-03T16:03:34","modified_gmt":"2023-10-03T16:03:34","slug":"week-6-and-7-ss2-first-term-mathematics-notes","status":"publish","type":"post","link":"https:\/\/ecolebooks.com\/nigeria\/posts\/week-6-and-7-ss2-first-term-mathematics-notes\/","title":{"rendered":"Week 6 and 7 &#8211; SS2 First Term Mathematics Notes"},"content":{"rendered":"<p>\u00a0<strong>WEEK SIX<br \/>\n<\/strong><strong>REVIEW OF FIRST HALF TERM LESSONS<br \/>\n<\/strong><br \/>\n\u00a0<br \/>\n\u00a0<strong>WEEK SEVEN<br \/>\n<\/strong><strong>TOPIC:  SIMULTANEOUS EQUATIONS<br \/>\n<\/strong><strong>CONTENT<br \/>\n<\/strong><\/p>\n<ul>\n<li>Solving Simultaneous Equations Using Elimination and Substitution Method\n<\/li>\n<li>Solving Equations Involving Fractions.\n<\/li>\n<li>Word problems.\n<\/li>\n<\/ul>\n<p><strong>SIMULTANEOUS LINEAR EQUATIONS<\/strong><br \/>\n\t\t<strong>Methods of solving Simultaneous equation<br \/>\n<\/strong>i.\u00a0\u00a0\u00a0\u00a0Elimination method<br \/>\nii.\u00a0\u00a0\u00a0\u00a0Substitution method<br \/>\niii.\u00a0\u00a0\u00a0\u00a0Graphical method <\/p>\n<p>\u00a0<strong>ELIMINATION METHOD<br \/>\n<\/strong>One of the unknowns with the same coefficient in the two equations is eliminated by subtracting or adding the two equations. Then the answer of the first unknown is substituted into either of the equations to get the second unknown.<strong><br \/>\n\t\t\t<\/strong><br \/>\n\u00a0<strong>Example<br \/>\n<\/strong>Solve for x and y in the equations 2x + 5y   = 1  and 3x \u2013 2y   =   30<br \/>\n<strong>Solution<br \/>\n<\/strong>To eliminate x multiply equation 1 by 3 and equation 2 by 2<strong><br \/>\n\t\t\t\t<\/strong>\u00a0\u00a0\u00a0\u00a02x + 5y = 1\u00a0\u00a0\u00a0\u00a0\u2026\u2026\u2026.\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0eqn 1 (x (3)<br \/>\n\u00a0\u00a0\u00a0\u00a03x \u2013 2y = 30 \u2026\u2026\u2026\u2026\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0eqn 2 (x (2)\u00a0\u00a0\u00a0\u00a0<br \/>\nResulting into,<br \/>\n\u00a0\u00a0\u00a0\u00a06x + 15 y = 3\u00a0\u00a0\u00a0\u00a0\u2026\u2026\u2026.\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0eqn 3<br \/>\n\u00a0\u00a0\u00a0\u00a06x \u2013 4y = 60   \u00a0\u00a0\u00a0\u00a0\u2026\u2026\u2026..\u00a0\u00a0\u00a0\u00a0eqn 4<br \/>\n\u00a0\u00a0\u00a0\u00a0Subtract eqn 3 from eqn 4<br \/>\n     6x \u2013 6x + 15y \u2013 (- 4y) = 3 \u2013 60<br \/>\n<img decoding=\"async\" align=\"left\" src=\"https:\/\/ecolebooks.com\/nigeria\/wp-content\/uploads\/9jalessonsimages\/100323_1600_Week6and71.png\" alt=\"\"\/><img decoding=\"async\" align=\"left\" src=\"https:\/\/ecolebooks.com\/nigeria\/wp-content\/uploads\/9jalessonsimages\/100323_1600_Week6and72.png\" alt=\"\"\/>\u00a0\u00a0\u00a0\u00a019y =   -57   3<br \/>\n<img decoding=\"async\" align=\"left\" src=\"https:\/\/ecolebooks.com\/nigeria\/wp-content\/uploads\/9jalessonsimages\/100323_1600_Week6and73.png\" alt=\"\"\/><img decoding=\"async\" align=\"left\" src=\"https:\/\/ecolebooks.com\/nigeria\/wp-content\/uploads\/9jalessonsimages\/100323_1600_Week6and74.png\" alt=\"\"\/>\u00a0\u00a0\u00a0\u00a0 19        19<br \/>\n\u00a0\u00a0\u00a0\u00a0<strong>y = -3<br \/>\n<\/strong><strong>\u00a0\u00a0\u00a0\u00a0<\/strong>Substitute y = &#8211; 3 into eqn 1<br \/>\n\u00a0\u00a0\u00a0\u00a02x + 5 (-3) = 1<br \/>\n\u00a0\u00a0\u00a0\u00a02x = 1 + 15<br \/>\n\u00a0\u00a0\u00a0\u00a02x  =16<br \/>\n\t\t\u00a0\u00a0\u00a0\u00a0 2         2<br \/>\n\u00a0\u00a0\u00a0\u00a0x  =  8<br \/>\n\u00a0\u00a0\u00a0\u00a0\\  y =  -3  and  x = 8<\/p>\n<p>\u00a0<strong>Evaluation<br \/>\n<\/strong>Using elimination method to solve the simultaneous equations.<br \/>\n1.  5x \u2013 4y = 38 and x + 3y = 22<br \/>\n2. 2c-3d= -4 and 4c-3d= -14<\/p>\n<p>\u00a0<strong>SUBSTITUTION METHOD<br \/>\n<\/strong>One of the unknowns (preferably the one having 1 has its coefficient) is made the subject of the formula in one of the equations and substituted into the other equation to obtain the value of the first  unknown which is then substituted into either of the equations to get the second unknown.<\/p>\n<p>\u00a0<strong>Example: <\/strong>Solve the simultaneous equation 2x + 5y = 1 and 3x \u2013 2y = 30<br \/>\n<strong>Solution<\/strong><br \/>\n\t\t2x + 5y   = 1\u2026\u2026\u2026\u2026\u2026. eq 1<br \/>\n3x \u2013 2y   = 30 \u2026\u2026\u2026\u2026.. eq 2<br \/>\nMake x the subject in eqn 1<br \/>\n2x   =   1 \u2013 5y<br \/>\n\t\t 2             2<br \/>\nx =   1 \u2013 5y   \u2026\u2026\u2026\u2026    eqn 3<br \/>\n           2<br \/>\nSubstitute eq3 into eqn 2<br \/>\n3 (1-5y)   &#8211;   2y = 30<br \/>\n      2<br \/>\nMultiple through by 2 or find the LCM and cross multiply.<br \/>\n3 \u2013 15y  &#8211;  4y  = 30<br \/>\n         2<br \/>\n3 \u2013 15y \u2013 4y = 60<br \/>\n3 \u2013 19y = 60<br \/>\n-19y = 60 \u2013 3<br \/>\n<img decoding=\"async\" align=\"left\" src=\"https:\/\/ecolebooks.com\/nigeria\/wp-content\/uploads\/9jalessonsimages\/100323_1600_Week6and75.png\" alt=\"\"\/><img decoding=\"async\" align=\"left\" src=\"https:\/\/ecolebooks.com\/nigeria\/wp-content\/uploads\/9jalessonsimages\/100323_1600_Week6and76.png\" alt=\"\"\/>-19y  = 57   3<br \/>\n\t\t\t<img decoding=\"async\" align=\"left\" src=\"https:\/\/ecolebooks.com\/nigeria\/wp-content\/uploads\/9jalessonsimages\/100323_1600_Week6and77.png\" alt=\"\"\/><img decoding=\"async\" align=\"left\" src=\"https:\/\/ecolebooks.com\/nigeria\/wp-content\/uploads\/9jalessonsimages\/100323_1600_Week6and78.png\" alt=\"\"\/> -19       -19<br \/>\ny  =  &#8211; 3<br \/>\nSubstitute y = -3 into eq 3<br \/>\nx  =1 \u2013 5y<br \/>\n\t\t           2<br \/>\nx   =   1 \u2013 5 (-3)=  1 + 15    =   16<br \/>\n\t\t\u00a0\u00a0\u00a0\u00a0   2\u00a0\u00a0\u00a0\u00a0          2\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 2<br \/>\n x =   8<strong><br \/>\n\t\t\t\t<\/strong>\\ x = 8, y = -3<br \/>\n<strong>Evaluation<br \/>\n<\/strong>Solve for x and y in the equations<br \/>\n1. x + 2y = 10 and 4x + 3y = 20<br \/>\n2. 4x-y=8 and 5x+y=19<\/p>\n<p>\u00a0<strong>SIMULTANEOUS EQUATIONS INVOLVING FRACTIONS<br \/>\n<\/strong><strong>Example<br \/>\n<\/strong>1. Solve the following equations simultaneously<br \/>\n\u00a0\u00a0\u00a0\u00a02 &#8211;   1 =  3    and         4  +   3    =   16<br \/>\n\u00a0\u00a0\u00a0\u00a0x      y                             x       y\u00a0\u00a0\u00a0\u00a0<br \/>\n<strong>Solution<br \/>\n<\/strong>\u00a0\u00a0\u00a0\u00a02   &#8211;    1    =    3<br \/>\n\u00a0\u00a0\u00a0\u00a0x        y\u00a0\u00a0\u00a0\u00a0<br \/>\n\u00a0\u00a0\u00a0\u00a04   +\u00a0\u00a0\u00a0\u00a03     =    16<br \/>\n\u00a0\u00a0\u00a0\u00a0x\u00a0\u00a0\u00a0\u00a0y<br \/>\nInstead of using x and y as the unknown, let the unknown be (<sup>1<\/sup>\/<sub>x<\/sub>) an (<sup>1<\/sup>\/<sub>y<\/sub>).<br \/>\n2(<sup>1<\/sup>\/<sub>x<\/sub>) &#8211;  (<sup>1<\/sup>\/<sub>y<\/sub>)  =  3\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u2026\u2026\u2026\u2026\u2026.\u00a0\u00a0\u00a0\u00a0eqn 1<br \/>\n4 (<sup>1<\/sup>\/<sub>x<\/sub>) &#8211; 3 (<sup>1<\/sup>\/<sub>y<\/sub>)   =  16\u00a0\u00a0\u00a0\u00a0\u2026\u2026\u2026\u2026\u2026\u00a0\u00a0\u00a0\u00a0eqn 2<br \/>\nUsing elimination method, multiply equation 1 by 2 to eliminate x.<br \/>\n4(<sup>1<\/sup>\/<sub>x<\/sub>) \u2013 2(<sup>1<\/sup>\/<sub>y<\/sub>)  =   6\u00a0\u00a0\u00a0\u00a0\u2026\u2026\u2026\u2026\u2026..     eqn 3<br \/>\n4 (<sup>1<\/sup>\/<sub>x<\/sub>) + 3(<sup>1<\/sup>\/<sub>y<\/sub>) =   16\u00a0\u00a0\u00a0\u00a0\u2026\u2026\u2026\u2026\u2026.\u00a0\u00a0\u00a0\u00a0eqn 4<br \/>\n<img decoding=\"async\" align=\"left\" src=\"https:\/\/ecolebooks.com\/nigeria\/wp-content\/uploads\/9jalessonsimages\/100323_1600_Week6and79.png\" alt=\"\"\/>\u00a0\u00a0\u00a0\u00a0-5 (<sup>1<\/sup>\/<sub>y<\/sub>)   =   -10<br \/>\n\t\t<img decoding=\"async\" align=\"left\" src=\"https:\/\/ecolebooks.com\/nigeria\/wp-content\/uploads\/9jalessonsimages\/100323_1600_Week6and710.png\" alt=\"\"\/>\u00a0\u00a0\u00a0\u00a0   -5\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 -5<br \/>\n\u00a0\u00a0\u00a0\u00a01   =    2<br \/>\n\u00a0\u00a0\u00a0\u00a0y<br \/>\n\u00a0\u00a0\u00a0\u00a0<strong>\\   y = \u00bd<br \/>\n<\/strong>Substitute (1\/y) = 2   into eqn 1<br \/>\n2 (<sup>1<\/sup>\/<sub>x<\/sub>) \u2013 (<sup>1<\/sup>\/<sub>y<\/sub>) = 3<br \/>\n2 (<sup>1<\/sup>\/<sub>x<\/sub>) \u2013 (2) = 3<br \/>\n2(<sup>1<\/sup>\/<sub>x<\/sub>) = 3 + 2<br \/>\n2 (<sup>1<\/sup>\/<sub>x<\/sub>) =   5<br \/>\n1     =       5<br \/>\nx\u00a0\u00a0\u00a0\u00a0   2<br \/>\n\\ x   = <sup>2<\/sup>\/<sub>5<\/sub><br \/>\n\t\t\t\\ y = \u00bd,  x  =  <sup>2<\/sup>\/<sub>5<\/sub><br \/>\n\t\t\t<strong>Evaluation<br \/>\n<\/strong>I. Solve for x and y simultaneously,      II. Solve the pair of equations for x and y<br \/>\nx  +y    =    1                                                 respectively.<br \/>\n             2      2                                                                  2x<sup>-1<\/sup> \u2013 3y<sup>-1<\/sup> = 4<br \/>\nx   &#8211;   y  =   1\u00bd                                                    4x<sup>-1<\/sup> + y<sup>-1<\/sup> = 1<br \/>\n2        6<br \/>\n<strong>FURTHER EXAMPLES<br \/>\n<\/strong>Solve for x and y simultaneously: 2x \u2013 3y + 2 = x + 2y \u2013 5 = 3x + y.<br \/>\n<strong>Solutions<br \/>\n<\/strong>\u00a0\u00a0\u00a0\u00a02x \u2013 3y + 2 = x + 2y \u2013 5 = 3x + y<br \/>\n\u00a0\u00a0\u00a0\u00a0Form two equations out of the question<br \/>\n\u00a0\u00a0\u00a0\u00a02x \u2013 3y + 2 = 3x + y<br \/>\n\u00a0\u00a0\u00a0\u00a0x + 2y \u2013 5 = 3x + y<br \/>\n\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0OR<br \/>\n\u00a0\u00a0\u00a0\u00a02x \u2013 3y + 2 = x + 2y \u2013 5 &#8212;&#8212;&#8212;&#8212;- eq 1<br \/>\n\u00a0\u00a0\u00a0\u00a0x + 2y \u2013 5 = 3x + y       &#8212;&#8212;&#8212;&#8212;&#8211; eq 2<br \/>\nRearrange the equations to put the unknown on one side and the constant at the other side.<br \/>\n\u00a0\u00a0\u00a0\u00a02x \u2013 3y \u2013 x \u2013 2y = &#8211; 5 \u2013 2<br \/>\n\u00a0\u00a0\u00a0\u00a02x \u2013 x \u2013 3y \u2013 2y = -7<br \/>\n\u00a0\u00a0\u00a0\u00a0x \u2013 5y = -7 &#8212;&#8212;&#8212;&#8212;&#8212;- eq 3<br \/>\n\u00a0\u00a0\u00a0\u00a0From eqn 2<br \/>\n\u00a0\u00a0\u00a0\u00a0x \u2013 3x + 2y \u2013 y \u2013 5<br \/>\n\u00a0\u00a0\u00a0\u00a0&#8211; 2x + y = 5 &#8212;&#8212;&#8212;&#8212;- eq 4<br \/>\nUsing substitution method solve eq 3 &amp; 4<br \/>\n\u00a0\u00a0\u00a0\u00a0x \u2013 5y = -7 &#8212;&#8212;&#8212;&#8212;&#8212;- eq 3<br \/>\n\u00a0\u00a0\u00a0\u00a0-2x + y = 5 &#8212;&#8212;&#8212;&#8212;&#8212; eq 4<br \/>\n\u00a0\u00a0\u00a0\u00a0Make y the subject in eq 4.<br \/>\n\u00a0\u00a0\u00a0\u00a0y = 5 + 2x &#8212;&#8212;&#8212;&#8212;&#8212; eq 5<br \/>\n\u00a0\u00a0\u00a0\u00a0Substitute eqn 5 into eqn 3.<br \/>\n\u00a0\u00a0\u00a0\u00a0x \u2013 5 (5 + 2x) = -7<br \/>\n\u00a0\u00a0\u00a0\u00a0x \u2013 25 \u2013 10x = -7<br \/>\n\u00a0\u00a0\u00a0\u00a0-9x \u2013 25 = -7<br \/>\n\u00a0\u00a0\u00a0\u00a0-9x = -7 + 25<br \/>\n\u00a0\u00a0\u00a0\u00a0-9x = 18<br \/>\n                 x = 18<br \/>\n\t\t\u00a0\u00a0\u00a0\u00a0          -9<br \/>\nX = -2<br \/>\nSubstitute x = &#8211; 2 into eqn 5<br \/>\ny = 5 + 2x<br \/>\ny = 5 + 2(-2)<br \/>\ny = 5 \u2013 4<br \/>\ny = 1<br \/>\n\\ x = -2, y = 1<br \/>\n<strong>Example<br \/>\n<\/strong>Solve the equations<br \/>\n              5<sup>x \u2013 y\/2<\/sup>  = 1                81<sup>x<\/sup>  =  27 <sup>3x -y<\/sup><br \/>\n\t\t\u00a0\u00a0\u00a0\u00a0                                   9<br \/>\n<strong>Solution<br \/>\n<\/strong>\u00a0\u00a0\u00a0\u00a05<sup>x \u2013 y\/2<\/sup>  = 1             &#8212;&#8212;&#8212;&#8211; eq 1<br \/>\n81<sup>x<\/sup>=  27<sup>3x -y<\/sup>        &#8212;&#8212;&#8212;- eq 2<br \/>\n\u00a0\u00a0\u00a0\u00a0  9<br \/>\nFrom eq 1 (using the law of indices)<br \/>\n5<sup>x \u2013 y\/2<\/sup>  = 5<sup>0<\/sup><br \/>\n\t\t x \u2013 y\/2 = 0<br \/>\n2x \u2013 y = 0 &#8212;&#8212;&#8212;&#8212; eq 3<br \/>\nFrom eq 2.<br \/>\n81<sup>x<\/sup>=  27<sup>3x -y<\/sup><br \/>\n\t\t\u00a0\u00a0\u00a0\u00a0 9<br \/>\n3 <sup>4x<\/sup>    =   3 <sup>3(3x-y)<\/sup><br \/>\n\t\t3 <sup>2 <\/sup><br \/>\n\t\t              3 <sup>4x-2<\/sup>  =  3 <sup>3(3x-y)<\/sup><br \/>\n\t\tBy comparison<br \/>\n\u00a0\u00a0\u00a0\u00a04x \u2013 2 = 9x \u2013 3y<br \/>\n\u00a0\u00a0\u00a0\u00a04x \u2013 9x + 3y = 2<br \/>\n\u00a0\u00a0\u00a0\u00a0&#8211; 5x + 3y =2 &#8212;&#8212;&#8212; eq 4<br \/>\nSolve equation 3 and 4 simultaneously<br \/>\n\u00a0\u00a0\u00a0\u00a02x \u2013 y = 0 &#8212;&#8212;&#8212; eq 3<br \/>\n\u00a0\u00a0\u00a0\u00a0-5x + 3y =2 &#8212;&#8212;&#8212;- eq 4<br \/>\n\u00a0\u00a0\u00a0\u00a0Using elimination method: multiply equation 3 by 3<br \/>\n\u00a0\u00a0\u00a0\u00a06x \u2013 3y = 0     &#8212;&#8212;&#8211; eq 3<br \/>\n\u00a0\u00a0\u00a0\u00a0-5x + 3y = 2 &#8212;&#8212;&#8212;- eq 4<br \/>\neq 3 + eq 4<br \/>\nx = 2<br \/>\nSubstitute x = 2 into eq 3<br \/>\n2x \u2013 y = 0<br \/>\n2 (2) \u2013 y = 0<br \/>\n4 \u2013 y = 0<br \/>\n4 = 0+y<br \/>\n4 = y<br \/>\n<strong>\\ x = 2, y=4<br \/>\n<\/strong><br \/>\n\u00a0<strong>WORD   PROBLEMS LEADING TO SIMULTANEOUS EQUATIONS<br \/>\n<\/strong><strong>Examples<br \/>\n<\/strong>1.Seven cups and eight plates cost N1750, eight cups and seven plates cost N1700. Calculate the cost of a cup and a plate<br \/>\n<strong>solution<br \/>\n\t\t\t\t<\/strong>\u00a0\u00a0\u00a0\u00a0Let   a cup be x and plate be y<br \/>\n\u00a0\u00a0\u00a0\u00a07x + 8y = 1750 &#8212;&#8212;&#8212;&#8212;&#8211; eq 1<br \/>\n\u00a0\u00a0\u00a0\u00a08x + 7y = 1700 &#8212;&#8212;&#8212;&#8212;&#8211; eq 2<\/p>\n<p>\u00a0\u00a0\u00a0\u00a0\u00a0Multiply eq 1 by 8 and eq 2 by 7 to eliminate x (cups).<br \/>\n\u00a0\u00a0\u00a0\u00a056x + 64y = 14000 &#8212;&#8212;&#8212;- eq 3<br \/>\n\u00a0\u00a0\u00a0\u00a056x + 49y = 11900 &#8212;&#8212;&#8212;- eq 4<br \/>\n\u00a0\u00a0\u00a0\u00a0Subtracting eq 4 from eq 3<br \/>\n15y  =  2100<br \/>\n                              y =   2100<br \/>\n\t\t                                         15<br \/>\n                               Y = 140<br \/>\n\u00a0\u00a0\u00a0\u00a0Substitute y = 140 into eq 2<br \/>\n\u00a0\u00a0\u00a0\u00a08x + 7y = 1700<br \/>\n\u00a0\u00a0\u00a0\u00a08x + 7 (140) = 1700<br \/>\n\u00a0\u00a0\u00a0\u00a08x + 980 = 1700<br \/>\n\u00a0\u00a0\u00a0\u00a08x = 1700 \u2013 980<br \/>\n               8x = 720<br \/>\n                  x = 720<br \/>\n                          8<br \/>\n                   x =  90<br \/>\n\u00a0\u00a0\u00a0\u00a0\\ Each cup cost N90 and each plate cost N140 <\/p>\n<p>\u00a02. Find a two digit number such that two times the tens digit is three less than thrice the unit digit and 4 times the number is 99 greater than the number obtained by reversing the digit.<br \/>\n<strong>Solution<br \/>\n<\/strong>Let the two digit number be  ab, where a is the tens digit and b is the unit digit<br \/>\n          From the first statement,<br \/>\n2a  + 3 = 3b<br \/>\n                                 2a \u2013 3b = -3 \u2026\u2026\u2026\u2026.eq1<br \/>\n          From the second statement,<br \/>\n                                  4(10a + b) \u2013 99 = 10b + a<br \/>\n                                  40a + 4b \u2013 99 = 10b + a<br \/>\n                                  40a \u2013 a + 4b \u2013 10b = 99<br \/>\n                                   39a \u2013 6b = 99<br \/>\n           Dividing through by 3<br \/>\n                                    13a \u2013 2b = 33 \u2026\u2026\u2026\u2026.eq2<br \/>\nSolving both equations simultaneously,<br \/>\n                                     a = 3 , b = 3<br \/>\nHence, the two digit number is 33<br \/>\n<strong>EVALUATION<br \/>\n<\/strong>1.The  sum  of  two  numbers  is  110  and  their  difference  is  20. Find the two numbers.          \u00a0\u00a0\u00a0\u00a0<br \/>\n2.A pen  a  ruler  cost  #30.If  the  pen  costs  #8  more  than  the  ruler, how  much  does  each  item  cost ?                                                                                                                                 <\/p>\n<p>\u00a0<strong>GENERAL EVALUATION AND REVISION   QUESTION<\/strong><br \/>\n\t\t1. Solve the   following simultaneous equation:  3(2x \u2013 y) = x + y + 5 &amp;  5(3x  &#8211;  2y) = 2 (x \u2013y) + 1<br \/>\n2. Five years ago, a father was 3 times as old as his son. Now, their combined ages amount to 110years. How old are they?<br \/>\n3. A  doctor  and  three  nurses  in  a  hospital  together  earn  #255 000 per  month, while  three  doctors  and  eight  nurses  together  earn  #720  000   per  month. Calculate (a) how much a doctor earns per month.  (b) How much  a  nurse  earns  per  month.<br \/>\n4.  Solve   simultaneously,   2<sup>x + 2y <\/sup>= 1; 3<sup>2x+y <\/sup>= 27<br \/>\n5.  Solve:  2x \u2013 2y + 5 = 3x \u2013 4y + 2 = -1                                                                       <\/p>\n<p>\u00a0<strong>WEEKEND ASSIGNMENT<br \/>\n<\/strong>1. If (x-y) log<sub>10<\/sub>6  = log<sub>10<\/sub> 216 and 2 <sup>x+y<\/sup> =32 , calculate the values of x and y <strong><br \/>\n\t\t\t<\/strong>a. x=1 , y=4        b. x= 4 , y =1       c. x=-4 , y= 1      d. x=4, y= -1<br \/>\n2. The point of intersection of the lines 3x- 2y =-12 and x + 2y = 4 is \u2026<br \/>\na. (5, 0)               b. (3, 4)               c. (-2, 5)              d. (-2, 3)<br \/>\n3. Find the value of (x &#8211; y), if 2x + 2y =16 and 8x \u2013 2y = 44   a. 2    b. 4    c. 5     d. 6<br \/>\n4. If 5 <sup>(p +2q)<\/sup> =5 and 4 <sup>(p+3q)<\/sup> =16, the value of  3<sup>(p+q)<\/sup>  is \u2026..     a.0   b. -1   c.2    d. 1<br \/>\n5.  Given 4x \u2013 3y = 11 evaluate y<sup>2<\/sup> \u2013 3x<br \/>\n\t\t7x \u2013 4y    23                     3                  a.   -2      b. 3       c. -3        d. 2<\/p>\n<p>\u00a0<strong>THEORY<br \/>\n<\/strong>1. Given that 2 <sup>1- x\/y<\/sup>  = 1\/32, find x in terms of y, and hence solve the simultaneous equations<br \/>\n2x + 3y \u2013 30 = 0 and        2<sup>1- x\/y<\/sup>  = 1\/32 (WAEC)<br \/>\n2. A number is made up of two digits. The sum of the digits is 11. If the digits are interchanged, the original number is increased by 9. Find the original number. (WAEC)<\/p>\n<p>\u00a0<strong>Reading assignment<br \/>\n<\/strong>Essential Mathematics for SSS2, pages 55-59, exercise 5.2<br \/>\n<strong><br \/>\n\t\t\t<\/strong>\u00a0<\/p>\n","protected":false},"excerpt":{"rendered":"<p>\u00a0WEEK SIX REVIEW OF FIRST HALF TERM LESSONS \u00a0 \u00a0WEEK SEVEN TOPIC: SIMULTANEOUS EQUATIONS CONTENT&#8230;<\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[1,233],"tags":[],"class_list":["post-2895","post","type-post","status-publish","format-standard","hentry","category-posts","category-first-term-ss2-mathematics"],"aioseo_notices":[],"_links":{"self":[{"href":"https:\/\/ecolebooks.com\/nigeria\/wp-json\/wp\/v2\/posts\/2895","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/ecolebooks.com\/nigeria\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/ecolebooks.com\/nigeria\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/ecolebooks.com\/nigeria\/wp-json\/wp\/v2\/users\/1"}],"replies":[{"embeddable":true,"href":"https:\/\/ecolebooks.com\/nigeria\/wp-json\/wp\/v2\/comments?post=2895"}],"version-history":[{"count":1,"href":"https:\/\/ecolebooks.com\/nigeria\/wp-json\/wp\/v2\/posts\/2895\/revisions"}],"predecessor-version":[{"id":2896,"href":"https:\/\/ecolebooks.com\/nigeria\/wp-json\/wp\/v2\/posts\/2895\/revisions\/2896"}],"wp:attachment":[{"href":"https:\/\/ecolebooks.com\/nigeria\/wp-json\/wp\/v2\/media?parent=2895"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/ecolebooks.com\/nigeria\/wp-json\/wp\/v2\/categories?post=2895"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/ecolebooks.com\/nigeria\/wp-json\/wp\/v2\/tags?post=2895"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}