Algebra InputsTrigonometry InputsCalculus InputsMatrix Inputs Basic algebra trigonometry calculus statistics matrices CharactersSolve for x, y x=3 y=2Steps Using SubstitutionSteps Using MatricesSteps Using EliminationView solution stepsSteps Using Substitution \left. \begin{array} { c } { 5 x - 6 y = 3 } \\ { 7 y = 2 x + 8 } \end{array} \right. Consider the second equation. Subtract 2x from both sides. 7y-2x=8 To solve a pair of equations using substitution, first solve one of the equations for one of the variables. Then substitute the result for that variable in the other equation. 5x-6y=3,-2x+7y=8 Choose one of the equations and solve it for x by isolating x on the left hand side of the equal sign. 5x-6y=3 Add 6y to both sides of the equation. 5x=6y+3 Divide both sides by 5. x=\frac{1}{5}\left(6y+3\right) Multiply \frac{1}{5} times 6y+3. x=\frac{6}{5}y+\frac{3}{5} Substitute \frac{6y+3}{5} for x in the other equation, -2x+7y=8. -2\left(\frac{6}{5}y+\frac{3}{5}\right)+7y=8 Multiply -2 times \frac{6y+3}{5}. -\frac{12}{5}y-\frac{6}{5}+7y=8 Add -\frac{12y}{5} to 7y. \frac{23}{5}y-\frac{6}{5}=8 Add \frac{6}{5} to both sides of the equation. \frac{23}{5}y=\frac{46}{5} Divide both sides of the equation by \frac{23}{5}, which is the same as multiplying both sides by the reciprocal of the fraction. y=2 Substitute 2 for y in x=\frac{6}{5}y+\frac{3}{5}. Because the resulting equation contains only one variable, you can solve for x directly. x=\frac{6}{5}\times 2+\frac{3}{5} Multiply \frac{6}{5} times 2. x=\frac{12+3}{5} Add \frac{3}{5} to \frac{12}{5} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible. x=3 The system is now solved. x=3,y=2 GraphQuizSimultaneous Equation5 problems similar to: \left. \begin{array} { c } { 5 x - 6 y = 3 } \\ { 7 y = 2 x + 8 } \end{array} \right.
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What is \displaystyle{\left[\begin{array}{c} -{3}{x}-{8}{y}={20}\\-{5}{x}+{y}={19}\end{array}\right]} ?https://socratic.org/questions/what-is-3x-8y-20-5x-y-19 \displaystyle{x}=-{4} \displaystyle{y}=-{1} Explanation: \displaystyle-{3}{x}-{8}{y}={20} ----------- (1) \displaystyle-{5}{x}+{y}={19} -------------(2) \displaystyle\times 8 ... We have matrix \displaystyle{A}={\left[\begin{array}{cc} {2}&{x}\\{y}&{3}\end{array}\right]} and \displaystyle{B}={\left[\begin{array}{cc} {y}&-{3}\\{5}&{2}{x}\end{array}\right]} , \displaystyle{A},{B}\in{M}{M}_{{{2},{2}}}{\left(\mathbb{C}\right)} ...https://socratic.org/questions/we-have-matrix-a-2-x-y-3-and-b-y-3-5-2x-a-binmm-2-2-cc-how-you-find-x-y-such-tha Douglas K. Sep 24, 2017 Given: \displaystyle{A}={\left[\begin{array}{cc} {2}&{x}\\{y}&{3}\end{array}\right]} and \displaystyle{B}={\left[\begin{array}{cc} {y}&-{3}\\{5}&{2}{x}\end{array}\right]} ... Find the equation of a line that satisfies the conditionshttps://math.stackexchange.com/q/2325529 When you got that the x-intercept of the line is 3, it already means that when x=3, y=z=0. So your '?' in (3, ?,?) are both 0. For another part, you got the direction ratios to be (0,b,b). ... Direction Field and Trajectorieshttps://math.stackexchange.com/q/1016358 In 1-D, it is called a direction field plot. In higher dimensions, it is called a phase portrait and these two sets of notes 1 and notes 2 provide a procedure for sketing it by hand. We look at ... Preserving orbits by multiplication with a non-vanishing functionhttps://math.stackexchange.com/questions/1013409/preserving-orbits-by-multiplication-with-a-non-vanishing-function Suppose that we have a solution x(t),y(t) to x'=f(x,y) y'=g(x,y). Define \tilde f(t) to satisfy the differential equation \tilde f'(t)=h(x(\tilde f(t)),y(\tilde f(t))). Notice that, ... Probability distribution for sum of RV'shttps://math.stackexchange.com/q/2340770 Corrections: Determinant of jacobian is 1 and limits of the integral should be 0 to z. You didn't use 0 < x < 1 \implies 0 < z-w<1.More Items
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CopyCopied to clipboard7y-2x=8 Consider the second equation. Subtract 2x from both sides.5x-6y=3,-2x+7y=8 To solve a pair of equations using substitution, first solve one of the equations for one of the variables. Then substitute the result for that variable in the other equation.5x-6y=3 Choose one of the equations and solve it for x by isolating x on the left hand side of the equal sign.5x=6y+3 Add 6y to both sides of the equation.x=\frac{1}{5}\left(6y+3\right) Divide both sides by 5.x=\frac{6}{5}y+\frac{3}{5} Multiply \frac{1}{5} times 6y+3.-2\left(\frac{6}{5}y+\frac{3}{5}\right)+7y=8 Substitute \frac{6y+3}{5} for x in the other equation, -2x+7y=8.-\frac{12}{5}y-\frac{6}{5}+7y=8 Multiply -2 times \frac{6y+3}{5}.\frac{23}{5}y-\frac{6}{5}=8 Add -\frac{12y}{5} to 7y.\frac{23}{5}y=\frac{46}{5} Add \frac{6}{5} to both sides of the equation.y=2 Divide both sides of the equation by \frac{23}{5}, which is the same as multiplying both sides by the reciprocal of the fraction.x=\frac{6}{5}\times 2+\frac{3}{5} Substitute 2 for y in x=\frac{6}{5}y+\frac{3}{5}. Because the resulting equation contains only one variable, you can solve for x directly.x=\frac{12+3}{5} Multiply \frac{6}{5} times 2.x=3 Add \frac{3}{5} to \frac{12}{5} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.x=3,y=2 The system is now solved.7y-2x=8 Consider the second equation. Subtract 2x from both sides.5x-6y=3,-2x+7y=8 Put the equations in standard form and then use matrices to solve the system of equations.\left(\begin{matrix}5&-6\\-2&7\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}3\\8\end{matrix}\right) Write the equations in matrix form.inverse(\left(\begin{matrix}5&-6\\-2&7\end{matrix}\right))\left(\begin{matrix}5&-6\\-2&7\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}5&-6\\-2&7\end{matrix}\right))\left(\begin{matrix}3\\8\end{matrix}\right) Left multiply the equation by the inverse matrix of \left(\begin{matrix}5&-6\\-2&7\end{matrix}\right).\left(\begin{matrix}1&0\\0&1\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}5&-6\\-2&7\end{matrix}\right))\left(\begin{matrix}3\\8\end{matrix}\right) The product of a matrix and its inverse is the identity matrix.\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}5&-6\\-2&7\end{matrix}\right))\left(\begin{matrix}3\\8\end{matrix}\right) Multiply the matrices on the left hand side of the equal sign.\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}\frac{7}{5\times 7-\left(-6\left(-2\right)\right)}&-\frac{-6}{5\times 7-\left(-6\left(-2\right)\right)}\\-\frac{-2}{5\times 7-\left(-6\left(-2\right)\right)}&\frac{5}{5\times 7-\left(-6\left(-2\right)\right)}\end{matrix}\right)\left(\begin{matrix}3\\8\end{matrix}\right) For the 2\times 2 matrix \left(\begin{matrix}a&b\\c&d\end{matrix}\right), the inverse matrix is \left(\begin{matrix}\frac{d}{ad-bc}&\frac{-b}{ad-bc}\\\frac{-c}{ad-bc}&\frac{a}{ad-bc}\end{matrix}\right), so the matrix equation can be rewritten as a matrix multiplication problem.\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}\frac{7}{23}&\frac{6}{23}\\\frac{2}{23}&\frac{5}{23}\end{matrix}\right)\left(\begin{matrix}3\\8\end{matrix}\right) Do the arithmetic.\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}\frac{7}{23}\times 3+\frac{6}{23}\times 8\\\frac{2}{23}\times 3+\frac{5}{23}\times 8\end{matrix}\right) Multiply the matrices.\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}3\\2\end{matrix}\right) Do the arithmetic.x=3,y=2 Extract the matrix elements x and y.7y-2x=8 Consider the second equation. Subtract 2x from both sides.5x-6y=3,-2x+7y=8 In order to solve by elimination, coefficients of one of the variables must be the same in both equations so that the variable will cancel out when one equation is subtracted from the other.-2\times 5x-2\left(-6\right)y=-2\times 3,5\left(-2\right)x+5\times 7y=5\times 8 To make 5x and -2x equal, multiply all terms on each side of the first equation by -2 and all terms on each side of the second by 5.-10x+12y=-6,-10x+35y=40 Simplify.-10x+10x+12y-35y=-6-40 Subtract -10x+35y=40 from -10x+12y=-6 by subtracting like terms on each side of the equal sign.12y-35y=-6-40 Add -10x to 10x. Terms -10x and 10x cancel out, leaving an equation with only one variable that can be solved.-23y=-6-40 Add 12y to -35y.-23y=-46 Add -6 to -40.y=2 Divide both sides by -23.-2x+7\times 2=8 Substitute 2 for y in -2x+7y=8. Because the resulting equation contains only one variable, you can solve for x directly.-2x+14=8 Multiply 7 times 2.-2x=-6 Subtract 14 from both sides of the equation.x=3 Divide both sides by -2.x=3,y=2 The system is now solved.
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Quadratic equation { x } ^ { 2 } - 4 x - 5 = 0Trigonometry 4 \sin \theta \cos \theta = 2 \sin \thetaLinear equation y = 3x + 4Arithmetic 699 * 533Matrix \left[ \begin{array} { l l } { 2 } & { 3 } \\ { 5 } & { 4 } \end{array} \right] \left[ \begin{array} { l l l } { 2 } & { 0 } & { 3 } \\ { -1 } & { 1 } & { 5 } \end{array} \right]Simultaneous equation \left. \begin{cases} { 8x+2y = 46 } \\ { 7x+3y = 47 } \end{cases} \right.Differentiation \frac { d } { d x } \frac { ( 3 x ^ { 2 } - 2 ) } { ( x - 5 ) }Integration \int _ { 0 } ^ { 1 } x e ^ { - x ^ { 2 } } d xLimits \lim _{x \rightarrow-3} \frac{x^{2}-9}{x^{2}+2 x-3}Back to top
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Steps Using SubstitutionSteps Using MatricesSteps Using EliminationView solution stepsSteps Using Substitution \left. \begin{array} { c } { 5 x - 6 y = 3 } \\ { 7 y = 2 x + 8 } \end{array} \right. Consider the second equation. Subtract 2x from both sides. 7y-2x=8 To solve a pair of equations using substitution, first solve one of the equations for one of the variables. Then substitute the result for that variable in the other equation. 5x-6y=3,-2x+7y=8 Choose one of the equations and solve it for x by isolating x on the left hand side of the equal sign. 5x-6y=3 Add 6y to both sides of the equation. 5x=6y+3 Divide both sides by 5. x=\frac{1}{5}\left(6y+3\right) Multiply \frac{1}{5} times 6y+3. x=\frac{6}{5}y+\frac{3}{5} Substitute \frac{6y+3}{5} for x in the other equation, -2x+7y=8. -2\left(\frac{6}{5}y+\frac{3}{5}\right)+7y=8 Multiply -2 times \frac{6y+3}{5}. -\frac{12}{5}y-\frac{6}{5}+7y=8 Add -\frac{12y}{5} to 7y. \frac{23}{5}y-\frac{6}{5}=8 Add \frac{6}{5} to both sides of the equation. \frac{23}{5}y=\frac{46}{5} Divide both sides of the equation by \frac{23}{5}, which is the same as multiplying both sides by the reciprocal of the fraction. y=2 Substitute 2 for y in x=\frac{6}{5}y+\frac{3}{5}. Because the resulting equation contains only one variable, you can solve for x directly. x=\frac{6}{5}\times 2+\frac{3}{5} Multiply \frac{6}{5} times 2. x=\frac{12+3}{5} Add \frac{3}{5} to \frac{12}{5} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible. x=3 The system is now solved. x=3,y=2 GraphQuizSimultaneous Equation5 problems similar to: \left. \begin{array} { c } { 5 x - 6 y = 3 } \\ { 7 y = 2 x + 8 } \end{array} \right.
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