Solving systems of equations Problem set B

Solve without a calculator. \(3x + 5y=17\) and \(4x - 3y=-16\).

Write each equation in the form \(ax^2+bx+c=0\).

1. (a) \((4x-1)(x+6)=15\)
2. (b) \(3x-\frac{8}{x}=5\)

Solve each system with your calculator. (Draw a graph for part (a) if you wish: the solution is where the lines meet.)

1. (a) \(7x + 3y=-5\), \(4x - 5y=14\)
2. (b) \(x=4y-3\), \(5x+2y=19\)

Solve using your calculator. \(3x - y + 2z=-7\), \(x + 4y - 3z=7\), \(2x + y + z=0\).

Rearrange into standard form, then solve using your calculator. \(2x + y=3z + 6\), \(-y + 2z=-4x + 1\), \(x + z - 2=-3y - 4\).

Solve using your calculator. \(2w + x + 3y - z=5\), \(w - 2x + y + 2z=-8\), \(-w + 3x + 2y + z=7\), \(3w + x - y + 2z=-6\).

Rearrange into standard form, then solve using your calculator. \(2w - x + 3y + z=6\), \(3w + x + 2z=y + 2\), \(w + 2x + 2y=3z - 5\), \(-2x + z=-2w - y + 4\).

Decide how many solutions each system has: none, one or infinitely many. Give a reason, then answer part (c).

1. (a) \(x + 2y = 5\) and \(3x + 6y = 20\)
2. (b) \(5x - 10y = 15\) and \(x - 2y = 3\)
3. (c) The system \(3x + ky = 7\) and \(12x + 20y = 5\) has no solution. Find \(k\).

A cafe sells coffees at one price and cakes at another. One table has 3 coffees and 2 cakes and pays $18.50. Another table has 5 coffees and 3 cakes and pays $29.45.

1. (a) Write two equations.
2. (b) Find the cost of 2 coffees and 4 cakes.

A saver's monthly deposits form an arithmetic sequence. The term \(u_{6}\) is 26 and the sum of the first 8 terms is 160. Find the first term \(u_1\) and the common difference \(d\).

A stack of logs is built layer by layer. The sum of the first 5 terms is 20 and the sum of the first 12 terms is 100.5. Find the first term \(u_1\) and the common difference \(d\).