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What are the solutions to the simultaneous equations of y plus 1 equals x and y equals x square -6x plus 5?
- y + 1 = x
- y = x² - 6x + 5
Rearrange {1} to make y the subject, substitute for y in the second and solve the quadratic:
y + 1 = x → y = x - 1
y = x² - 6x + 5
→ x - 1 = x² - 6x + 5
→ x² - 7x + 6 = 0
→ (x - 1)(x - 6) = 0
→ x = 1 or 6
→ x = 1: y = 1 - 1 = 0
& x = 6 → y = 6 - 1 = 6
→ the solutions are the points (1, 0) and (6, 5)
What else can I help you with?
What are the solutions to the equation y2 equals 169?
y2 = 169 Square root both sides: y = 13
What are the solutions to the equation x2 equals 81?
x2 = 81 Square root both sides:- x = +/- 9
What reasoning and explanations can be used when solving radical equations?
The basic method is the same as for other types of equations: you need to isolate the variable ("x", or whatever variable you need to solve for). In the case of radical equations, it often helps to square both sides of the equation, to get rid of the radical. You may need to rearrange the equation before squaring. It is important to note that when you do this (square both sides), the new equation may have solutions which are NOT part of the original equation. Such solutions are known as "extraneous" solutions. Here is a simple example (without radicals): x = 5 (has one solution, namely, 5) Squaring both sides: x squared = 25 (has two solutions, namely 5, and -5). To protect against this situation, make sure you check each "solution" of the modified equation against the original equation, and reject the solutions that don't satisfy it.
How do you solve x2 - 4x 41?
Equations have solutions. I'll assume that's x^2 - 4x + 41 = 0 since this website can't reproduce plus or equals signs in questions. That doesn't factor neatly. Using the quadratic formula, we find two imaginary solutions: 2 plus or minus i times the square root of 37x = 8.082762530298219ix = -4.082762530298219iwhere i is the imaginary square root of -1no its is x^2-4x=41
The square root of a sum equals the sum of the square roots?
false
What are the solutions to the simultaneous equations of x square plus y square plus 4x plus 6y minus 40 equals 0 and x minus y equals 10?
Rearrange the second equation as x = 10+y and then substitute it into the first equation which will create a quadratic equation in the form of: 2y2+30y+100 = 0 and when solved y = -10 or y = -5 Therefore the solutions are: x = 0, y = -10 and x = 5, y = -5
What are the solutions to the simultaneous equations of y equals -2x and x2 plus y2 equals 80?
If: y = -2x then y ^2 = 4x^2 If: x^2 + y^2 = 80 then x^2 +4x^2 = 80 So: 5x^2 = 80 Divide all terms by 5: x^2 = 16 Square root both sides: x = -4 or +4 By substitution into the original equation solutions are: (-4, 8) and (4, -8)
What math terms start with s?
Some of them out of many more are as follows:- Square Sphere Semi-circle Sector of a circle Segment of a circle Square root Simple percentages Solids and their nets Scatter graphs Simultaneous equations Solutions Straight line equations Slope of a line Subtraction Substitution Surface area Sine ratio of a right angle triangle Scalene triangle has no equal sides Sum
The perimeter of a rectangle is 26 feet and tge area is 30 square feet What is the width of the rectangle?
Write two simultaneous equations and solve them. One for the perimeter, one for the area.
How can you find out how many solutions an equation has?
By solving it. There is no single easy way to solve all equations; different types of equations required different methods. You have to learn separately how to solve equations with integer polynomials, rational equations (where polynomials can also appear in the denominator), equations with square roots and other roots, trigonometric equations, and others.Sometimes, the knowledge of a type of equations can help you quickly guess the number of solutions. Here are a few examples. An equation like:sin(x) = 0.5has an infinite number of solutions, because the sine function is periodic. An equation with a polynomial - well, in theory, you can factor a polynomial of degree "n" into "n" linear factors, meaning the polynomial can have "n" solutions. However, it may have multiple solutions, that is, some of the factors may be equal. Also, some of the solutions may be complex. A real polynomial of odd degree has at least one real solution.By solving it. There is no single easy way to solve all equations; different types of equations required different methods. You have to learn separately how to solve equations with integer polynomials, rational equations (where polynomials can also appear in the denominator), equations with square roots and other roots, trigonometric equations, and others.Sometimes, the knowledge of a type of equations can help you quickly guess the number of solutions. Here are a few examples. An equation like:sin(x) = 0.5has an infinite number of solutions, because the sine function is periodic. An equation with a polynomial - well, in theory, you can factor a polynomial of degree "n" into "n" linear factors, meaning the polynomial can have "n" solutions. However, it may have multiple solutions, that is, some of the factors may be equal. Also, some of the solutions may be complex. A real polynomial of odd degree has at least one real solution.By solving it. There is no single easy way to solve all equations; different types of equations required different methods. You have to learn separately how to solve equations with integer polynomials, rational equations (where polynomials can also appear in the denominator), equations with square roots and other roots, trigonometric equations, and others.Sometimes, the knowledge of a type of equations can help you quickly guess the number of solutions. Here are a few examples. An equation like:sin(x) = 0.5has an infinite number of solutions, because the sine function is periodic. An equation with a polynomial - well, in theory, you can factor a polynomial of degree "n" into "n" linear factors, meaning the polynomial can have "n" solutions. However, it may have multiple solutions, that is, some of the factors may be equal. Also, some of the solutions may be complex. A real polynomial of odd degree has at least one real solution.By solving it. There is no single easy way to solve all equations; different types of equations required different methods. You have to learn separately how to solve equations with integer polynomials, rational equations (where polynomials can also appear in the denominator), equations with square roots and other roots, trigonometric equations, and others.Sometimes, the knowledge of a type of equations can help you quickly guess the number of solutions. Here are a few examples. An equation like:sin(x) = 0.5has an infinite number of solutions, because the sine function is periodic. An equation with a polynomial - well, in theory, you can factor a polynomial of degree "n" into "n" linear factors, meaning the polynomial can have "n" solutions. However, it may have multiple solutions, that is, some of the factors may be equal. Also, some of the solutions may be complex. A real polynomial of odd degree has at least one real solution.
What are the roots of the equation x²-10x plus 25 equals 0?
Equations have 'solutions', not roots.Both of the solutions to this quadratic equation are x=5 .(The expression on the left side is a perfect square. So the graph of theexpression is a parabola whose nose rests on the x-axis at x=5, andboth x-intercepts are at the same point.)
Which of the numbers are solutions to the equation x2 equals 3?
If: x2 = 3 Then: x = square root of 3
What are the solutions to the simultaneous equations of 3x -5y equals 16 and xy equals 7 shown by completing the square?
If: 3x -5y = 16 then y = 0.6x -3.2 If: xy = 7 then x(0.6x -3.2) -7 = 0 Removing brackets: 0.6x^2 -3.2x -7 = 0 Dividing all terms by 0.6: x^2 -16/3x -35/3 = 0 Completing the square: (x -8/3)^2 -64/9 -35/3 = 0 => (x -8/3)^2 = 169/9 Square root both sides: x -8/3 = -13/3 or +13/3 Add 8/3 to both sides: x = -5/3 or x = 7 Solutions by substitution are: x = -5/3, y = -21/5 and x = 7, y = 1
What are the solutions to the equation x2 equals 81?
x2 = 81 Square root both sides:- x = +/- 9
What are the solutions to the equation y2 equals 169?
y2 = 169 Square root both sides: y = 13
What is x2 -2x plus 2 equals 0?
It is a quadratic equation with no real roots or real solutions. In the complex domain, the solutions are 1 +/- i where i is the imaginary square root of -1.
What is the perpendicular distance from the point 4 -2 to the line 2x -y -5 equals 0?
From the given information the perpendicular line will form an equation of 2y = -x and both simultaneous line equations will intersect each other at (2,-1) and so distance from (4, -2) to (2, -1) is the square root of 5 by using the distance formula.
