Systems of Linear Equations: math test solutions, Grade 9 – download the PDF

1 Checking pairs / 2 pts

\((4, -2)\): \(3(4) + 2(-2) = 8\) and \(4 - 3(-2) = 10\), both true, so it is a solution. (1 pt)

\((1, 5)\): \(3(1) + 2(5) = 13 \neq 8\), so it is not a solution. (1 pt)

2 Solve by graphing / 3 pts

Table: \(y = x - 2\) gives -2, -1, 0, 1, 2; \(y = -2x + 7\) gives 7, 5, 3, 1, -1. (1 pt)

The lines cross at \((3, 1)\), the only \(x\) where both rows are equal. (1 pt)

Check: \(3 - 2 = 1\) and \(-2(3) + 7 = 1\). The solution is \((3, 1)\). (1 pt)

3 Substitution / 3 pts

Substitute: \(5x + 2(-2x + 1) = 1\). (1 pt)

So \(5x - 4x + 2 = 1\), which gives \(x = -1\). (1 pt)

Then \(y = -2(-1) + 1 = 3\). Check: \(5(-1) + 2(3) = 1\). The solution is \((-1, 3)\). (1 pt)

4 Elimination / 3 pts

Multiply the first equation by 3 and the second by 2: \(12x + 15y = 36\) and \(12x - 14y = -80\). (1 pt)

Subtract: \(29y = 116\), so \(y = 4\). (1 pt)

Then \(4x + 20 = 12\), so \(x = -2\). Check: \(6(-2) - 7(4) = -40\). The solution is \((-2, 4)\). (1 pt)

5 How many solutions? / 2 pts

(a) The second equation gives \(-2y = -6x + 4\), so \(y = 3x - 2\): same line, infinitely many solutions. (1 pt)

(b) The second equation gives \(y = -x + 5\): slope -1 like the first line but intercept 5 instead of 1. The lines are parallel, so there is no solution. (1 pt)

6 Bike rental / 4 pts

(a) Shop A: \(c = 15 + 4h\). Shop B: \(c = 9 + 6h\). (1 pt)

(b) Set them equal: \(15 + 4h = 9 + 6h\), so \(6 = 2h\) and \(h = 3\). Then \(c = 15 + 4(3) = 27\). After 3 hours both shops charge 27 dollars. (2 pts)

(c) For \(h = 5\): shop A costs \(15 + 20 = 35\) dollars and shop B costs \(9 + 30 = 39\) dollars. Shop A is cheaper. (1 pt)

7 Inequalities / 3 pts

(a) The line \(x + y = 6\) is solid (because of \(\le\)); the line \(y = 2x - 3\) is dashed (because of \(\gt\)). (1 pt)

(b) \((1, 1)\): \(1 + 1 = 2 \le 6\) and \(1 \gt 2(1) - 3 = -1\), both true, so it is a solution. \((3, 2)\): \(5 \le 6\) is true but \(2 \gt 3\) is false, so it is not a solution. (1 pt)

(c) Solve \(x + (2x - 3) = 6\): \(3x = 9\), so \(x = 3\) and \(y = 3\). The lines meet at \((3, 3)\). (1 pt)