| Solved by pluggable solver: Solving a System of Linear Equations by Elimination/Addition |
Lets start with the given system of linear equations In order to solve for one variable, we must eliminate the other variable. So if we wanted to solve for y, we would have to eliminate x (or vice versa). So lets eliminate x. In order to do that, we need to have both x coefficients that are equal but have opposite signs (for instance 2 and -2 are equal but have opposite signs). This way they will add to zero. So to make the x coefficients equal but opposite, we need to multiply both x coefficients by some number to get them to an equal number. So if we wanted to get 11 and 2 to some equal number, we could try to get them to the LCM. Since the LCM of 11 and 2 is 22, we need to multiply both sides of the top equation by 2 and multiply both sides of the bottom equation by -11 like this: So after multiplying we get this: Notice how 22 and -22 add to zero (ie Now add the equations together. In order to add 2 equations, group like terms and combine them So after adding and canceling out the x terms we're left with: Now plug this answer into the top equation So our answer is which also looks like ( Notice if we graph the equations (if you need help with graphing, check out this solver) we get and we can see that the two equations intersect at ( |
| Solved by pluggable solver: SOLVE linear system by SUBSTITUTION |
| Solve: into another equation: Answer: |
| Solved by pluggable solver: Finding a distance between a point given by coordinates (x, y) and a line given by equation y=ax+b |
| We want to find the perpendicular distance between a point given by coordinates ( and a line given by equation First, let's draw a diagram of general situation with point P (xo, yo) and line L: y= a.x + b. The required distance is PC. (in the diagram below) Methodology We will first find the vertices of the triangle in order to get the side lengths and then by applying Sine Rule on right angle triangle PAB and PBC we will calculate the desired distance PC. Step1 Calculation of the vertices of triangle PAB: Draw a vertical line passing through the point 'P'. This line at point 'A'. The X coordinate of A(x1) will be same as 'A' we will use the fact that point 'A' lies on the given line 'L' and satisfies the equation of the line 'L' . Now, plug this Hence, Point (A)( Similarly, Draw a horizontal line passing through the point 'P'. This line at point 'B'. The Y coordinate of B(y2) will be same as B we will use the fact that point 'B' lies on the given line 'L' and satisfies the equation of the line 'L' . Now, plug this Hence, Point (B)( Now, we have all the vertices of the triangle PAB Step2 Calculation of the side lengths using distance formula: Hence, The side lengths PA, PB and AB are Step3 Apply Sine rule on common angle B in triangle PAB and triangle PBC. Both triangle PAB and triangle PBC are right angle triangle and points 'A', 'B' and 'C' lay on the given line L. PC is the required perpendicular distance of the point P (-8, -9) from line given lineL1: y=10*x+-9. For better understanding of this concept, look at the Lesson based on the above concept. Lesson |
| Solved by pluggable solver: Finding a distance between a point given by coordinates (x, y) and a line given by equation y=ax+b |
| We want to find the perpendicular distance between a point given by coordinates ( and a line given by equation First, let's draw a diagram of general situation with point P (xo, yo) and line L: y= a.x + b. The required distance is PC. (in the diagram below) Methodology We will first find the vertices of the triangle in order to get the side lengths and then by applying Sine Rule on right angle triangle PAB and PBC we will calculate the desired distance PC. Step1 Calculation of the vertices of triangle PAB: Draw a vertical line passing through the point 'P'. This line at point 'A'. The X coordinate of A(x1) will be same as 'A' we will use the fact that point 'A' lies on the given line 'L' and satisfies the equation of the line 'L' . Now, plug this Hence, Point (A)( Similarly, Draw a horizontal line passing through the point 'P'. This line at point 'B'. The Y coordinate of B(y2) will be same as B we will use the fact that point 'B' lies on the given line 'L' and satisfies the equation of the line 'L' . Now, plug this Hence, Point (B)( Now, we have all the vertices of the triangle PAB Step2 Calculation of the side lengths using distance formula: Hence, The side lengths PA, PB and AB are Step3 Apply Sine rule on common angle B in triangle PAB and triangle PBC. Both triangle PAB and triangle PBC are right angle triangle and points 'A', 'B' and 'C' lay on the given line L. PC is the required perpendicular distance of the point P (-9, -5) from line given lineL1: y=9*x+-21. For better understanding of this concept, look at the Lesson based on the above concept. Lesson |
| Solved by pluggable solver: Finding the Equation of a Line |
| First lets find the slope through the points ( So the slope is ------------------------------------------------ Now let's use the point-slope formula to find the equation of the line: ------Point-Slope Formula------ So lets use the Point-Slope Formula to find the equation of the line ------------------------------------------------------------------------------------------------------------ Answer: So the equation of the line which goes through the points ( The equation is now in Notice if we graph the equation Notice how the two points lie on the line. This graphically verifies our answer. |