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Q: You are given two equations which are both true and you are asked to solve for both x and y You plan to solve this set of equations by substituting part of one equation into the other so you end up?

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An equation has an equals sign ( = ). Equations assert the absolute equality of two expressions.

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I suggest that the simplest way is as follows:Assume the equation is of the form y = ax2 + bx + c.Substitute the coordinates of the three points to obtain three equations in a, b and c.Solve these three equations to find the values of a, b and c.

Without any equality signs the expessions given can't be considered to be equations and therefore have no solutions.

25+11=36: Let f and s represent the first and second numbers respectively. The statement of the problem yields two equations: f + s =36 and f = 3 + 2s. Substituting the function given in the second equation for f into the first equation yields 3 + 2s + s = 36, or (subtracting 3 from each side and merging the s terms, 3s = 33 or s = 11. Then f + 11 = 36 (substituting the value for s into the first equation), or f = 25.

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Many equations can be plotted as a line or other curve in the x-y plane. If you have two equations and you plot the curve (or line) for each one. If the curves (or lines) intersect at one or more points, then the (x,y) coordinates for the intersection points will also be the x and y values which satisfy both equations to be true at the same time.

An equation has an equals sign ( = ). Equations assert the absolute equality of two expressions.

There is only one equation that is given in the question and that equation is not a direct variation.

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CPUs, when given mathematical equations, apply the laws of mathematics to those equations. The equation a = a is true by the reflexive property of equality.

The elements and compounds to the right of the equations are called products.

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The answer will depend on the context. If the curve in question is a differentiable function then the gradient of the tangent is given by the derivative of the function. The gradient of the tangent at a given point can be evaluated by substituting the coordinate of the point and the equation of the tangent, though that point, is then given by the point-slope equation.

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