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###### Exercises

Exercise name | Free? |
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Exercises 1 Equations written in slope-intercept form follow a specific format. y=mx+b In this form, m represents the slope of the line and b represents the y-intercept.Identifying Slope and Intercept Since the given equation is not in slope-intercept form, let's rewrite it so that it will be easier to identify the slope and y-intercept. y+4=xLHS−4=RHS−4y=x−4 Below we highlight the slope m and y-intercept b. y=1x − 4⇔y=1x+(-4) The slope is 1. The y-intercept is -4, so the graph crosses the y-axis at the point (0,-4).Graphing the Equation A slope of 1 means that for every 1 unit we move in the positive horizontal direction, we move 1 unit in the positive vertical direction. m=11⇔runrise=11 To graph the equation, plot the y-intercept and then use the slope to find another point on the line. | |

Exercises 2 Equations written in slope-intercept form follow a specific format. y=mx+b In this form, m represents the slope of the line and b represents the y-intercept.Identifying Slope and Intercept Since the given equation is not in slope-intercept form, let's rewrite it so that it will be easier to identify the slope and y-intercept. 6x−y=−1LHS+1=RHS+16x−y+1=0LHS+y=RHS+y6x+1=yRearrange equationy=6x+1 Below we highlight the slope m and y-intercept b. y=6x+1 The slope is 6. The y-intercept is 1, so the graph crosses the y-axis at the point (0,1).Graphing the Equation A slope of 6 means that for every 1 unit we move in the positive horizontal direction, we move 6 units in the positive vertical direction. m=16⇔runrise=16 To graph the equation, plot the y-intercept and then use the slope to find another point on the line. | |

Exercises 3 Equations written in slope-intercept form follow a specific format. y=mx+b In this form, m represents the slope of the line and b represents the y-intercept.Identifying Slope and Intercept Since the given equation is not in slope-intercept form, let's rewrite it so that it will be easier to identify the slope and y-intercept. 4x+5y=20LHS−4x=RHS−4x5y=-4x+20LHS/5=RHS/5y=-54x+4 Below we highlight the slope m and y-intercept b. y=-54x+4 The slope is -54. The y-intercept is 4, so the graph crosses the y-axis at the point (0,4).Graphing the Equation A slope of -54 means that for every 5 units we move in the positive horizontal direction, we move 4 units in the negative vertical direction. m=-54⇔runrise=5-4 To graph the equation, plot the y-intercept and then use the slope to find another point on the line. | |

Exercises 4 Equations written in slope-intercept form follow a specific format. y=mx+b In this form, m represents the slope of the line and b represents the y-intercept.Identifying Slope and Intercept Since the given equation is not in slope-intercept form, let's rewrite it so that it will be easier to identify the slope and y-intercept. -2y+12=-3xLHS−12=RHS−12-2y=-3x−12LHS/-2=RHS/-2y=23x+6 Below we highlight the slope m and y-intercept b. y=23x+6 The slope is 23. The y-intercept is 6, so the graph crosses the y-axis at the point (0,6).Graphing the Equation A slope of 23 means that for every 2 units we move in the positive horizontal direction, we move 3 units in the positive vertical direction. m=23⇔runrise=23 To graph the equation, plot the y-intercept and then use the slope to find another point on the line. | |

Exercises 5 Inequalities can be solved in the same way as equations, by performing inverse operations on both sides until the variable is isolated. The only difference is that when you divide or multiply by a negative number, you must reverse the inequality sign. m+4>9LHS−4≤RHS−4m>5 This inequality tells us that all values greater than 5 will satisfy the inequality. Below we demonstrate the inequality by graphing the solution set on a number line. Notice that m can't equal 5, which we show with an open circle on the number line. | |

Exercises 6 Inequalities can be solved in the same way as equations, by performing inverse operations on both sides until the variable is isolated. The only difference is that when you divide or multiply by a negative number, you must reverse the inequality sign. 24≤-6tLHS/-6≥RHS/-6-8≥tRearrange inequalityt≤-8 This inequality tells us that all values less than or equal to -8 will satisfy the inequality. Below we demonstrate the inequality by graphing the solution set on a number line. Notice that t can equal -8, which we show with a closed circle on the number line. | |

Exercises 7 Inequalities can be solved in the same way as equations, by performing inverse operations on both sides until the variable is isolated. The only difference is that when you divide or multiply by a negative number, you must reverse the inequality sign. 2a−5≤13LHS+5≤RHS+52a≤18LHS/2≤RHS/2a≤9 This inequality tells us that all values less than or equal to 9 will satisfy the inequality. Below we demonstrate the inequality by graphing the solution set on a number line. Notice that a can equal 9, which we show with a closed circle on the number line. | |

Exercises 8 Inequalities can be solved in the same way as equations, by performing inverse operations on both sides until the variable is isolated. The only difference is that when you divide or multiply by a negative number, you must reverse the inequality sign. -5z+1<-14LHS−1<RHS−1-5z<-15LHS/3>RHS/3z>3 This inequality tells us that all values greater than 3 will satisfy the inequality. Below we demonstrate the inequality by graphing the solution set on a number line. Notice that z can't equal -3, which we show with an open circle on the number line. | |

Exercises 9 Inequalities can be solved in the same way as equations, by performing inverse operations on both sides until the variable is isolated. The only difference is that when you divide or multiply by a negative number, you must reverse the inequality sign. 4k−16<k+2LHS−2<RHS−24k−18<kLHS−4k<RHS−4k-18<-3kLHS/-3>RHS/-36>kRearrange inequalityk<6 This inequality tells us that all values less than 6 will satisfy the inequality. Below we demonstrate the inequality by graphing the solution set on a number line. Notice that k can't equal 6, which we show with an open circle on the number line. | |

Exercises 10 Inequalities can be solved in the same way as equations, by performing inverse operations on both sides until the variable is isolated. The only difference is that when you divide or multiply by a negative number, you must reverse the inequality sign. 7w+12≥2w−3LHS+3≥RHS+37w+15≥2wLHS−7w≥RHS−7w15≥-5wLHS/-5≤RHS/-5-3≤wRearrange inequalityw≥-3 This inequality tells us that all values greater than or equal to -3 will satisfy the inequality. Below we demonstrate the inequality by graphing the solution set on a number line. Notice that w can equal -3, which we show with a closed circle on the number line. | |

Exercises 11 We know that the value of both functions at x=a is y=b. We also know that they have different slopes. Let's plot the arbitrary point (a,b) on a coordinate plane and graph any two lines with different slopes that meet at (a,b).We can see that the point (a,b) is where the lines intersect. |

##### Other subchapters in Solving Systems of Linear Equations

- Mathematical Practices
- Solving Systems of Linear Equations by Graphing
- Solving Systems of Linear Equations by Substitution
- Solving Systems of Linear Equations by Elimination
- Solving Special Systems of Linear Equations
- Quiz
- Solving Equations by Graphing
- Graphing Linear Inequalities in Two Variables
- Systems of Linear Inequalities
- Chapter Review
- Chapter Test
- Cumulative Assessment