1) Three times a number, minus another number, is 19. The sum of the two numbers is 13. Find the two numbers.
Let the first number be \(x\) and the second number be \(y\). Then
\begin{align}3x-y & =19 \quad \enclose{circle}{\color{black}{1}}\\ x+y & =13 \quad \enclose{circle}{\color{black}{2}}\\ add \; eqn\;\enclose{circle}{\color{black}{1}}\; and \; eqn \; \enclose{circle}{\color{black}{2}} \; to\; eliminate \; y\\ (3x-y)+(x+y) & =19+13\\ 4x & =32\\ x & =8\\ then\; by\; substitution\; into\; \enclose{circle}{\color{black}{1}}\\ 3(8)-y & =19\\ -y & = -5\\ y & =5 \end{align}So \(x=8\), \(y=5\)
2) Three times a number plus two times a larger number is 28. And six times the smaller number plus two times the larger number is 40. Find the two numbers.
Let the smaller number be \(x\) and the larger number be \(y\). Then
\begin{align}3x+2y & =28 \quad \enclose{circle}{\color{black}{1}}\\ 6x+2y & =40 \quad \enclose{circle}{\color{black}{2}}\\ subtract \; eqn\; \enclose{circle}{\color{black}{2}}\; from \; eqn \; \enclose{circle}{\color{black}{1}} \; to\; eliminate \; y\\ (6x+2y)-(3x+2y) & =40-28\\ 3x & =12\\ x & =4\\ then\; by\; substitution\; into\; \enclose{circle}{\color{black}{1}}\\ 3(4)+2y & =28\\ 2y & = 16\\ y & =8 \end{align}So \(x=4\), \(y=8\)
3) Three times a number minus four times a smaller number is 2. The difference between the two numbers is 4. Find the two numbers.
Let the larger number be \(x\) and the smaller number be \(y\). Then
\begin{align} 2x-4y & =2 \quad \enclose{circle}{\color{black}{1}}\\ x-y & = 4 \quad \enclose{circle}{\color{black}{2}}\\ multiply\; equation \; \enclose{circle}{\color{black}{2}}\; by\; 2\\ to \; equate\; the \; coefficients \; of \; x\\ 2x-2y & =8 \quad \enclose{circle}{\color{black}{3}}\\ subtract\; eqn\; \enclose{circle}{\color{black}{1}}\; from\; eqn\; \enclose{circle}{\color{black}{3}}\\ to \; eliminate\; x\\ (2x-2y)-(2x-4y) & =8-2\\ 2y & =6\\ y & = 3\\ then\; by\; substitution\; into\; \enclose{circle}{\color{black}{2}}\\ x-3 & =4\\ x & =7 \end{align}So \(x=7\), \(y=3\)
4) In a game of Dodali, one team scored seven 'dods' and nine 'dalis' giving them a total of 51 points. The opposing team scored four 'dods' and eighteen 'dalis', giving them a total of 42 points. How many points are there for each 'dod' and for each 'dali'?
Let points per 'dod' be \(x\) and points per 'dali' be \(y\). Then
\begin{align} 7x+9y & =51 \quad \enclose{circle}{\color{black}{1}}\\ 4x+18y & = 42 \quad \enclose{circle}{\color{black}{2}}\\ multiply\; equation \; \enclose{circle}{\color{black}{1}} \; by\; 2\\ to \; equate\; the \; coefficients \; of \; y\\ 14x+18y & =102 \quad \enclose{circle}{\color{black}{3}}\\ subtract\; eqn\; \enclose{circle}{\color{black}{1}} \; from\; eqn\; \enclose{circle}{\color{black}{3}}\\ to \; eliminate\; y\\ (14x+18y)-(4x+18y) & =102-42\\ 10x & =60\\ x & = 6\\ then\; by\; substitution\; into\; \enclose{circle}{\color{black}{1}}\\ 7(6)+9y & = 51\\ 9y & = 51-42\\ y & =1 \end{align}So points per 'dod' is \(6\), and points per 'dali' is \(1\)
5) A farmer can buy four sheep and two cows for $2,800, or he can buy two sheep and four cows for $4,850. What is the price of each sheep and each cow?
Let cost per sheep be \(x\) and cost per cow be \(y\). Then
\begin{align} 4x+2y & = 2,800 \quad \enclose{circle}{\color{black}{1}}\\ 2x+4y & = 4,850 \quad \enclose{circle}{\color{black}{2}}\\ multiply\; equation \; \enclose{circle}{\color{black}{1}} \; by\; 2\\ to \; equate\; the \; coefficients \; of \; y\\ 8x+4y & =5,600 \quad \enclose{circle}{\color{black}{3}}\\ subtract\; eqn\; \enclose{circle}{\color{black}{1}} \; from\; eqn\; \enclose{circle}{\color{black}{3}}\\ to \; eliminate\; y\\ (8x+4y)-(2x+4y) & =5,600-4,850\\ 6x & =750\\ x & = 125\\ then\; by\; substitution\; into\; \enclose{circle}{\color{black}{1}}\\ 4(125)+2y & = 2,800\\ 2y & = 2,800-500\\ 2y & =2,300\\ y & =1,150 \end{align}So the cost of a sheep is \($125\), and the cost of a cow is \($1,150\)
6) Twenty five red bricks and seven white bricks laid end-to-end measure 3.80 metres. Twelve white bricks and fourteen red bricks laid end-to-end measure 3.34 metres. What is the length of a red brick?
Let the length of a red brick be \(x\) and the length of a white brick be \(y\). Then
\begin{align} 25x+7y & = 380 \quad \enclose{circle}{\color{black}{1}}\\ 14x+12y & = 334 \quad \enclose{circle}{\color{black}{2}}\\ multiply\; equation \; \enclose{circle}{\color{black}{1}} \; by\; 12\\ and \; multiply \; equation \; \enclose{circle}{\color{black}{2}} \; by \; 7\\ to \; equate\; the \; coefficients \; of \; y\\ 12(25x)+12(7y) & = 12(380)\\ 300x+84y & =4560 \quad \enclose{circle}{\color{black}{3}}\\ 7(14x)+7(12y) & =7(334)\\ 98x+84y & =2338 \quad \enclose{circle}{\color{black}{4}}\\ subtract\; eqn\; \enclose{circle}{\color{black}{4}}\; from\; eqn\; \enclose{circle}{\color{black}{3}}\\ to \; eliminate\; y\\ (300x+84y)-(98x+84y) & =4,560-2,338\\ 202x & =2222\\ x & = 11\\ \end{align}So the length of a red brick is \(11cm\)
7) The cost of five apples and seven oranges is $3.33. The cost of three apples and four oranges is $1.94. Find the cost of one apple, and the cost of one orange.
Let the cost of an apple be \(x\), and the cost of an orange be \(y\). Then
\begin{align}5x+7y & = 333 \quad \enclose{circle}{\color{black}{1}}\\ 3x+4y & = 194 \quad \enclose{circle}{\color{black}{2}}\\ multiply \; eqn \; \enclose{circle}{\color{black}{1}} \; by \; 3,\; and\; eqn \; \enclose{circle}{\color{black}{2}}\; by\; 5\\ to \; equate \; the \; x\; coefficients\\ 15x+21y & =999 \quad \enclose{circle}{\color{black}{3}}\\ 15x+20y & = 970 \quad \enclose{circle}{\color{black}{4}}\\ Subtract \; \enclose{circle}{\color{black}{4}} \; from \; \enclose{circle}{\color{black}{3}} \; to\; eliminate \; x\\ (15x+21y)-(15x+20y) & =999-970\\ y & = 29\\ then\; by\; substitution\; into\; \enclose{circle}{\color{black}{1}}\\ 5x+7(29) & =333\\ 5x & = 333-203\\ 5x & =130\\ x & = 26 \end{align}So one apple costs 26 cents, and one orange costs 29 cents.
8) Eight bags of flour and six bags of sugar weigh 12kg altogether. Twelve bags of flour reduced by the weight of two bags of sugar weighs 9.2 kg. What is the weight of a bag of sugar?
Let the weight of a bag of flour be \(x\), and the weight of a bag of sugar be \(y\).
\begin{align}8x+6y & = 12,000 \quad \enclose{circle}{\color{black}{1}}\\ 12x-2y & = 9,200 \quad \enclose{circle}{\color{black}{2}}\\ multiply \; eqn \; \enclose{circle}{\color{black}{2}}\: by\; 3\\ to \; equate \; the \; y\; coefficients\\ 36x-6y & =27,600 \quad \enclose{circle}{\color{black}{3}}\\ Add \; \enclose{circle}{\color{black}{3}} \; to \; \enclose{circle}{\color{black}{1}} \; to\; eliminate \; y\\ (8x+6y)+(36x-6y) & =12,000+27,600\\ 44x & = 39,600\\ x & =900\\ then\; by\; substitution\; into\; \enclose{circle}{\color{black}{1}}\\ 8(900)+6y & =12,000\\ 6y & = 12,000 - 7,200\\ 6y & =4,800\\ y & = 800 \end{align}So each bag of sugar weighs \(800gm\)