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 Transcript Example 18 Solve the following pair of equations by reducing them to a pair of linear equations 5/(𝑥 −1) 1/(𝑦 −2) = 2 6/(𝑥 −1) – 3/(𝑦 −2) = 1 5/(𝑥 − 1) 1/(𝑦 − 2) = 2 6/(𝑥 − 1) – 3/(𝑦 − 2) = 1 So, our equations become 5u v = 2 6u – 3v = 1 Thus, our equations are 5u v = 2 (3) 6u – 3v = 1 (4) From (3) 5u v = 2 v = 2 Transcript Ex 36, 1 Solve the following pairs of equations by reducing them to a pair of linear equations (i) 1/2𝑥 1/3𝑦 = 2 1/3𝑥 1/2𝑦 = 13/6 1/2𝑥 1/3𝑦 = 2 1/3𝑥 1/2𝑦 = 13/6 Let 1/𝑥 = u 1/𝑦 = v So, our equations become 1/2 u 1/3 v = 2 (3𝑢 2𝑣)/(2 × 3) = 2 3u 2v = 12 1/3 u 1/2 v = 13/6 (2𝑢 3𝑣)/(2 × 3) = 13/6 2u 3v = 13 Our equations(1/2) x (3/2) (x 1) 1/4 = 5 (1/2) x (3/2) x 3/2 1/4 = 5 2x = 5 3/2 1/4 2x = /4 6/4 1/4 2x = 15/4 x = 15/8

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5/x-1 1/y-2=2 6/x-1-3/y-2=1 by reducing method-Graph y=1/5x2 Rewrite in slopeintercept form Tap for more steps The slopeintercept form is , where is the slope and is the yintercept Write in form Tap for more steps Reorder terms Remove parentheses Use the slopeintercept form to find the slope and yintercept(2/3)x1=5 Simple and best practice solution for (2/3)x1=5 equation Check how easy it is, and learn it for the future Our solution is simple, and easy to understand, so don`t hesitate to use it as a solution of your homework If it's not what You are looking for type in the equation solver your own equation and let us solve it

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0 votes 1 answerSolve for x/2 y/5=1 for y Answer y= 5/2x 5 Step 1 The answer is in slopeintercept form where m is the slope and b is the yintercept when x=0 at point (0,b) In this example, the answer reveals that m=5/2 and b=5 Step 2 To get this equation from , use the following stepsSimple and best practice solution for Y35=21 equation Check how easy it is, and learn it for the future Our solution is simple, and easy to understand, so don`t hesitate to use it as a solution of your homework If it's not what You are looking for type in the equation solver your own equation and let us solve it

5 2 2 5 1 1 = 3 1 1 We've found the nonzero eigenvector x 2 = 1 1 with corresponding eigenvalue 2 = 3 Check that this also gives a solution by plugging y 1 = e3t and y 2 = 3et back into the di erential equations Notice that we've found two independent solutions x 1 and x 2 More is true, you can see that x 1 is actually perpendicular to13x 12y = 136 (ii) 2√(x) 3√(y) = 2 ; color(red)(x=6) Start from the given equation 2/3x1/6=1/2x5/6 We will use 6 on both sides of the equation to get rid of the fractions Multiply both sides by 6 6(2/3x1/6)=6(1/2x5/6) (12x)/36/6=(6x)/230/6 Simplify 4x1=3x5 4x3x=51 color(red)(x=6) Checking the answer x=6 into the original equation 2/3x1/6=1/2x5/6 2/3(6)1/6=1/2(6)5/6 12/31/6=6/25/6 23/6=23/6

4/√x 9/√y = 1You can put this solution on YOUR website!Steps Using the Quadratic Formula y = \frac { x ^ { 2 } 1 } { x ^ { 2 } 5 x 6 } y = x 2 − 5 x − 6 x 2 − 1 Variable x cannot be equal to any of the values 1,6 since division by zero is not defined Multiply both sides of the equation by \left (x6\right)\left (x1\right) Variable x cannot be equal to any of the values − 1, 6

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First, the expression needs to be rewritten as x 2 a x b x 6 To find a and b, set up a system to be solved ab=5 ab=1\times 6=6 a b = − 5 a b = 1 × 6 = 6 Since ab is positive, a and b have the same sign Since ab is negative, a and b are both negative List all such integer pairs that give product 66x 1 3y 2 = 1 (v) 7x 2yxy = 5 ;Y5=1/6(x2) Simple and best practice solution for y5=1/6(x2) equation Check how easy it is, and learn it for the future Our solution is simple, and easy to understand, so don`t hesitate to use it as a solution of your homework If it's not what You are looking for type in the equation solver your own equation and let us solve it

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Solve for x 2/(x2)1/5=6/(x5) Subtract from both sides of the equation Expand using the FOIL Method Tap for more steps Apply the distributive property Apply the distributive property Apply the distributive property Simplify and combine like terms Tap for more stepsWhat is 2 1/6 x 3 2/5 2 1/6 * 3 2/5 = 13/6 * 17/5 = 221/30 Answer 7 11/30 obuong3@aolcomShare It On Facebook Twitter Email 1 Answer 1 vote answered by Md samim (950k points) selected by sforrest072 Best answer The correct answer is ← Prev

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2x 4y = 5xy (vii) 10x y 2xGraph y5=1/2*(x2) Move all terms not containing to the right side of the equation Tap for more steps Add to both sides of the equation Add and Rewrite in slopeintercept form Tap for more steps The slopeintercept form is , where is the slope and is the yintercept Reorder termsSolve the following pairs of equations by reducing them to a pair of linear equations (i) 1/2x 1/3y = 2;

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 Transcript Example 17 Solve the pair of equations 2/𝑥 3/𝑦=13 5/𝑥−4/𝑦=−2 2/𝑥 3/𝑦=13 5/𝑥−4/𝑦=−2 So, our equations become 2u 3v = 13 5u – 4v = –2 Hence, our equations are 2u 3v = 13 (3) 5u – 4v = – 2 (4) From (3) 2u 3v = 13 2u = 13 – 3V u = (13 − 3𝑣)/2 Putting value of u (4) 5u – 4v = 2 5((13 − 3𝑣)/2)−4𝑣=−2 MultiplyingSteps for Solving Linear Equation y=2x1 y = 2 x − 1 Swap sides so that all variable terms are on the left hand side Swap sides so that all variable terms are on the left hand side 2x1=y 2 x − 1 = y Add 1 to both sides Add 1 to both sidesCombine x1 2 x 1 2 and 1 y2 3 1 y 2 3 Change the sign of the exponent by rewriting the base as its reciprocal Apply the product rule to y2 3 x1 2 y 2 3 x 1 2 Multiply the exponents in (y2 3)6 ( y 2 3) 6 Tap for more steps Apply the power rule and multiply exponents, ( a m) n = a

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The second term of the sum is equal to negative two The bottom of the rational expression is a sum that comprises 2 terms The first term of the sum is equal to a power The base is x The exponent is two The second term of the sum is equal to negative x x plus negative two fraction line x to the power two plus negative x3 That's it Step3 Rewrite the polynomial splitting the middle term using the two factors found in step 2 above, 1 and 2 x2 1x 2x 2 Step4 Add up the first 2 terms, pulling out like factors x • (x1) Add up the last 2 terms, pulling out common factors 2 • (x1) Step5 Add up the four terms of step 4 Ex 36, 1 Solve the following pairs of equations by reducing them to a pair of linear equations (vii) 10/(𝑥 𝑦) 2/(𝑥 − 𝑦) = 4 15/(𝑥 𝑦) − 5/(𝑥 − 𝑦) = −2 10/(𝑥 𝑦) 2/(𝑥 − 𝑦) = 4 15/(𝑥 𝑦) – 5/(𝑥 − 𝑦) = –2 So, our equations become 10u 2v = 4 15u – 5v = –2 Now, we

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51 Adding fractions which have a common denominator Combine the numerators together, put the sum or difference over the common denominator then reduce to lowest terms if possible (2x3) (5) 2x 2 ———————————— = —————— 6 6The given equations can be written in the matrix form as `(2,1),(3,5)("x"),("y")=(5),(3)` By 2R 2, we get, `(2,1),(6,10)("x"),("y")=(5),(6)` By R 2 x = 2, y = 9 You have a simple system of linear equations and can use the substitution method So take one of the equations and solve for x We are going to use the second equation, because the coefficient of x is already 1 x 2/5y = 8/5 x = 8/5 2/5y Now let's substitute the x in our first equation 2/5x 1/5y = 1 2/5*(8/52/5y) 1/5y = 1 16/25 4/25y 1/5y = 1 1/25y = 9/25 y

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 Explanation 1 XXXX(x − 3)2 (y − 5)2 = 25 2 XXXXx2 (y − 1)2 = 100 Expanding the square for 1 3 XXXXx2 −6x 9 y2 − 10y 25 = 25 Simplifying 4 XXXXx2 −6x y2 −10y = − 9 Expanding the square for 2 5 XXXXx2 y2 − 2y 1 = 100Solve for X and Y `5/(X1) 2/(Y−1) = 1/2, 10/(X1) 2/(Y−1) = 5/2, Where X ≠ 1, Y ≠ 1` CBSE CBSE (English Medium) Class 10 Question Papers 6 Textbook Solutions Important Solutions 3111 Question Bank Solutions 334 Concept Notes & Videos 262 Time3x 4y = 23 (iv) 5x 1 1y 2 = 2 ;

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Y=3 (x5) (x2) Simple and best practice solution for y=3 (x5) (x2) equation Check how easy it is, and learn it for the future Our solution is simple, and easy to understand, so don`t hesitate to use it as a solution of your homework If it's not what You are looking for type in the equation solver your own equation and let us solve it0 votes 1 answer 2431 views around the world You can reuse this answer Creative Commons License

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 1/7x 1/6y =3 1/2x1/3y =5 pair of linear equations in two variables; Solve the system of equations by using the method of cross multiplication 5/xy 2/x− y 1 = 0, 15/xy 7/x− y – 10 = 0 asked Jun 23 in Linear Equations by Hailley (334k points) linear equations in two variables;Weekly Subscription $199 USD per week until cancelled Monthly Subscription $699 USD per month until cancelled Annual Subscription $3999 USD per year until cancelled

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8x 7yxy = 15 (vi) 6x 3y = 6xy ;Simple and best practice solution for 2/x5/2=1/2 equation Check how easy it is, and learn it for the future Our solution is simple, and easy to understand, so 5/x11/y2=2 6/x13/y2=1 solve this with reducing method mam Maths Pair of Linear Equations in Two Variables

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 Solve each of the following systems of equations by the method of cross multiplication 2/x 3/y = 13 5/x 4/y = 2 where x ≠ 0 and y ≠ 0 asked Apr 27 in Linear Equations by Gargi01 (242k points) pair of linear equations in two variables;EXERCISE 36 CLASS 10 MATHS CHAPTER 3LINEAR EQUATIONS IN TWO VARIABLES NCERT Solutions For Class 10 Maths Chapter 3 Linear Equations In Two Variables Ex 36 given here are prepared by the subject experts Visit now to download solutions in PDF for freeSteps for Solving Linear Equation x = 2 y 5 x = 2 y − 5 Swap sides so that all variable terms are on the left hand side Swap sides so that all variable terms are on the left hand side 2y5=x 2 y − 5 = x Add 5 to both sides Add 5 to both sides

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 Explanation When you have an equation with fractions, you can rid of the denominators straight away and then carry on to solve the equation Multiply each term by the LCM of the denominators so that you can cancel them In this case it is 12 2 3 x 1 = 5 6x 3 4 × LCM ( in this case LCM = 12) 12× 2x 3 12 ×1 = 12 × 5x 6 12 ×3 4Tangents to the curve y = x3 − 3x2 −7x6 cut off on the negative semi axis OX a line segment half that on the positive semi axis OY is Let the point of tangency be (x_1,y_1) Then the equation of tangent is \displaystyle\frac {yy_1} {xx_1}=3x_1^26x_17 When y=0, \displaystyle x=x_1\frac {y_1} {3x_1^26x_17}For example 1/2 and 2/4 are equivalent, y/(y1) 2 and (y 2 y)/(y1) 3 are equivalent as well To calculate equivalent fraction , multiply the Numerator of each fraction, by its respective Multiplier

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Learn with Tiger how to do x2/31/6=5/6 fractions in a clear and easy way Equivalent Fractions,Least Common Denominator, Reducing (Simplifying) Fractions Tiger Algebra SolverClick here👆to get an answer to your question ️ Solve the following pairs of equations by reducing them to a pair of linear equations(i) 12x 13y = 2 ; Transcript Ex 36, 1 Solve the following pairs of equations by reducing them to a pair of linear equations (v) (7𝑥 − 2𝑦)/𝑥𝑦 = 5 (8𝑥 7𝑦)/𝑥𝑦 = 15 Given (7𝑥 − 2𝑦)/𝑥𝑦 = 5 (7𝑥 )/𝑥𝑦 − (2𝑦 )/𝑥𝑦 = 5 (7 )/𝑦 −(2 )/𝑥 = 5 (−𝟐 )/𝒙 (𝟕 )/𝒚 = 5 (8𝑥 7𝑦)/𝑥𝑦 = 15 (8𝑥 )/𝑥𝑦 (7𝑦 )/𝑥𝑦 = 15 (8

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 5/x11/y2=2 eq 1 6/x13/y2=1 eq 2 let 1/x1=u, 1/y2=v multiply eq 1 by 3, we get 15u=3v=6 eq 3 add eq 2 and 3 21u=7 u=1/3 x1=31/3x 1/2y = 13/6 (ii) 2/√x 3/√y = 2;For example 1/2 and 2/4 are equivalent, y/(y1) 2 and (y 2 y)/(y1) 3 are equivalent as well To calculate equivalent fraction , multiply the Numerator of each fraction, by its respective Multiplier

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Assignment No

Solve the following pairs of equations by reducing them to a pair of linear equations 5/ (x1) 1/y2 = 2, 6/ (x1) 3/ (y2) = 1 Mathematics ShaalaacomSolve for x x15=2 Move all terms not containing to the right side of the equation Tap for more steps Subtract from both sides of the equation Subtract from Remove the absolute value term This creates a on the right side of the equation because Set up the positive portion of the solution4√(x) 9√(y) = 1 (iii) 4x 3y = 14 ;

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