Question

1. vocabulary how can you determine the number of solutions of a system by looking at the equations? lesson 5-1

1. how many solutions does the system of equations have? explain. lesson 5-1

2x - 9y = -5
4x - 6y = 2

1. graph the system of equations and find the solution. lesson 5-2

y = 2x - 1
y = \\(\frac{1}{2}\\)x + 2

1. one equation in a system is y = -3x + 7. which equation gives the system no solution? lesson 5-1

a y = -3x + 7 c y = -3x + 5
b y = 3x + 5 d y = \\(\frac{1}{3}\\)x - 7

1. finn bought 12 movie tickets. student tickets cost $4, and adult tickets cost $8. finn spent a total of $60. write and graph a system of equations to find the number of student and adult tickets finn bought. lesson 5-2

1. what value of m gives the system infinitely many solutions? lesson 5-1

-x + 4y = 32
y = mx + 8

how well did you do on the mid-topic checkpoint? fill in the stars.
topic 5 analyze and solve systems of linear equations

Explanation:

Question 2:

Step1: Recall solution types for linear systems

A system of linear equations \(a_1x + b_1y = c_1\) and \(a_2x + b_2y = c_2\) has:

• No solution if \(\frac{a_1}{a_2}=\frac{b_1}{b_2}

eq\frac{c_1}{c_2}\)

• One solution if \(\frac{a_1}{a_2}

eq\frac{b_1}{b_2}\)

• Infinitely many solutions if \(\frac{a_1}{a_2}=\frac{b_1}{b_2}=\frac{c_1}{c_2}\)

Given system: \(2x - 9y = -5\) (Equation 1) and \(4x - 6y = 2\) (Equation 2)

Step2: Check ratios of coefficients

Calculate \(\frac{a_1}{a_2}=\frac{2}{4}=\frac{1}{2}\), \(\frac{b_1}{b_2}=\frac{-9}{-6}=\frac{3}{2}\)

Since \(\frac{1}{2}
eq\frac{3}{2}\), the slopes are different (for linear equations in slope - intercept form, slope \(m =-\frac{a}{b}\)). So the lines intersect at one point.

Step1: Graph \(y = 2x-1\)

• For \(y = 2x - 1\), when \(x = 0\), \(y=-1\) (y - intercept \((0, - 1)\)). When \(y = 0\), \(2x-1 = 0\Rightarrow x=\frac{1}{2}\) (x - intercept \((\frac{1}{2},0)\)). Plot these points and draw the line.

Step2: Graph \(y=\frac{1}{2}x + 2\)

• For \(y=\frac{1}{2}x + 2\), when \(x = 0\), \(y = 2\) (y - intercept \((0,2)\)). When \(y = 0\), \(\frac{1}{2}x+2=0\Rightarrow x=-4\) (x - intercept \((-4,0)\)). Plot these points and draw the line.

Step3: Find intersection (solution)

To find the solution algebraically (to confirm), set \(2x-1=\frac{1}{2}x + 2\)

Subtract \(\frac{1}{2}x\) from both sides: \(2x-\frac{1}{2}x-1=\frac{1}{2}x-\frac{1}{2}x + 2\)

Simplify: \(\frac{3}{2}x-1 = 2\)

Add 1 to both sides: \(\frac{3}{2}x=3\)

Multiply both sides by \(\frac{2}{3}\): \(x = 2\)

Substitute \(x = 2\) into \(y = 2x-1\): \(y=2(2)-1=3\)

Step1: Recall condition for no solution in linear systems (slope - intercept form)

A system of linear equations \(y = m_1x + b_1\) and \(y = m_2x + b_2\) has no solution if \(m_1=m_2\) (same slope) and \(b_1
eq b_2\) (different y - intercepts), because parallel lines (same slope, different y - intercepts) never intersect.

Given equation: \(y=-3x + 7\) (slope \(m_1=-3\), y - intercept \(b_1 = 7\))

Step2: Analyze each option

• Option A: \(y=-3x + 7\). Here \(m=-3\), \(b = 7\). Same slope and same y - intercept, so infinitely many solutions.
• Option B: \(y = 3x+5\). Slope \(m = 3

eq - 3\), so lines intersect (one solution).

• Option C: \(y=-3x + 5\). Slope \(m=-3\) (same as given equation), y - intercept \(b = 5

eq7\). So parallel lines, no solution.

• Option D: \(y=\frac{1}{3}x-7\). Slope \(m=\frac{1}{3}

eq - 3\), so lines intersect (one solution).

Answer:

The system has one solution because the ratios of the coefficients of \(x\) (\(\frac{2}{4}=\frac{1}{2}\)) and the coefficients of \(y\) (\(\frac{-9}{-6}=\frac{3}{2}\)) are not equal, meaning the lines have different slopes and will intersect at exactly one point.

Question 3: