<\/p>\n
Linear Equation in Two Variables<\/strong><\/h2>\nThe linear equation in two variables in the form of ax + by + c = 0 generally represents a line <\/span>on the coordinate axes (graph)<\/span>. The easiest way to represent the equation on the graph is using the substitution method, i.e first assuming a value of a variable and finding out the value of the other.<\/span><\/p>\n<\/span><\/p>\nExample:<\/b><\/p>\n
3x + 2y – 6 = 0\u00a0<\/span><\/p>\nThe above equation has two variables i.e, x and y.<\/span><\/p>\nGraphically we can represent this equation by substituting the variables to zero.<\/span><\/p>\nThe value of x when y=0 is, we can write the equation as –<\/span><\/p>\n3x + 2(0) = 6<\/span><\/p>\n\u21d2 x = 2<\/span><\/p>\nWe show this solution on the graph as (2,0)<\/span><\/p>\nTherefore the value of y when x = 0 is,<\/span><\/p>\n3(0) + 2y = 6<\/span><\/p>\n\u21d2 y = 3<\/span><\/p>\nWe show this solution on the graph as (0, 3)<\/span><\/p>\n<\/span><\/p>\n![\"\"](\"https:\/\/eistudymaterial.s3.amazonaws.com\/Pair-of-Linear-Equation-in-Two-Variables-Mindspark-01.png\")
<\/span><\/p>\n<\/span><\/h2>\nTypes of Solution of Pair of Linear Equation with Two Variables-<\/strong><\/h2>\n<\/span><\/p>\n![\"\"](\"https:\/\/eistudymaterial.s3.amazonaws.com\/Pair-of-Linear-Equation-in-Two-Variables-Mindspark-02-300x122.png\")
<\/span><\/p>\n <\/p>\n
Unique Solution –<\/strong> When the lines represented on the coordinate axes by the equations intersect each other, then we say that the pair of equations has a unique solution.<\/p>\nNo Solution –<\/strong> When the lines represented on the coordinate axes by the equations do not intersect and are parallel to each other, then we say that the pair of equations has no solution.<\/p>\nInfinite Solutions –<\/strong> When the lines represented on the coordinate axes by the equations are coincident i.e, they both overlap each other ( form the same line) then we say that the pair of equations have infinite solutions.<\/p>\n\n
Methods To Solve Pair Of Linear Equation In Two Variables-<\/strong><\/h2>\n1. Graphical Method :<\/h3>\n\n- First, we plot the two-variable equations on the coordinate axes(graph).<\/li>\n
- Then we see where both the lines are intersecting.<\/li>\n
- The intersecting point is the solution.<\/li>\n<\/ul>\n
Example:<\/strong> Find the value of x and y by solving the given equations by graphical method.
x – y + 4 = 0 – (1) and x + y – 10 = 0 – (2)<\/p>\nFirst, we solve each equation separately and plot it. While plotting it on the axes, we find a point of intersection between the two equations. The point of intersection will be the solution.<\/p>\n
As we can see in the figure given below, the point of intersection is (3,7).<\/p>\n
Therefore the solution is x = 3 and y = 7.<\/p>\n
<\/p>\n
![\"\"](\"https:\/\/eistudymaterial.s3.amazonaws.com\/prl-300x194.png\")
<\/p>\n
<\/p>\n
2. Substitution Method :<\/h3>\n\n- First, solve both the equations for one variable.<\/li>\n
- Then by using the first variable, find the value of the second variable.<\/li>\n<\/ul>\n
Example <\/b>: <\/b>Solve the pair of equations by substitution method:\u00a0<\/span><\/p>\n5x \u2013 8y = 2\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0– (1)<\/span><\/p>\n\u00a0x + 4y = 3\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 – (2)<\/span><\/p>\nWe can\u00a0 pick either of the equations and write one variable in terms of the other, here taking equation (2) first:<\/span><\/p>\nSo, we can write\u00a0 x + 4y = 3 as\u00a0 \u00a0 x = 3 – 4y\u00a0 \u00a0 – (3)<\/span><\/p>\nAfter substituting the value of x in equation (1), we get –<\/span><\/p>\n5 (3 – 4y) + 8y = 2<\/span><\/p>\n\u21d2 15 – 20y + 8 = 2<\/span><\/p>\n\u21d2 23 – 20y = 2<\/span><\/p>\n\u21d2 – 20y = -21<\/span><\/p>\n\u21d2 y = 21\/20<\/span><\/p>\nNow we substitute the value of y in equation (3),<\/span><\/p>\nx = 3 – 4 (21\/20)<\/span><\/p>\n\u21d2 x = 3 – 21\/5<\/span><\/p>\n\u21d2 x = (15-21)\/5<\/span><\/p>\n\u21d2 x = – 6\/5<\/span><\/p>\nTherefore the solution is x = -6\/5 and y = 21\/20.<\/span><\/p>\n<\/span><\/p>\n3. Cross Multiplication Method :<\/span><\/h3>\n\n- Consider pair of linear equation in two variables as a_1\\text{x} + b_1\\text{y} +c_1=0 \\text{ and } a_2\\text{x}=b_2\\text{y} +c_2<\/span><\/span>.<\/span><\/span><\/li>\n
- Now write the variables, coefficients and the constant in the following way-<\/span><\/li>\n<\/ul>\n
![\"\"](\"https:\/\/eistudymaterial.s3.amazonaws.com\/prl-2-300x136.png\")
<\/span><\/p>\n\n- \u00a0Now we write the equation by cross multiplying and subtracting the product.<\/span><\/li>\n<\/ul>\n
![\"\"](\"https:\/\/eistudymaterial.s3.amazonaws.com\/prl-3-300x69.png\")
<\/span><\/p>\n\n- \u00a0\u00a0Now after solving the equations for x and y we get :\u00a0<\/span><\/li>\n<\/ul>\n
![\"\"](\"https:\/\/eistudymaterial.s3.amazonaws.com\/prl-5-300x61.png\")
<\/span><\/p>\n<\/b><\/p>\n
Example <\/b>:<\/b> Solve the following pair of linear equation by cross multiplication method\u00a0<\/span><\/p>\n3x – 5y = – 2\u00a0 and y – 2x = 6<\/span><\/p>\nThe given equations can be written as:\u00a0<\/span><\/p>\n3x – 5y +2\u00a0 = 0 \u00a0 \u00a0 – (1)<\/span><\/p>\n-2x + y – 6 = 0 \u00a0 \u00a0 – (2)<\/span><\/p>\nNow by using cross multiplication we get\u00a0<\/span><\/p>\nx=\\frac{b_{1} \\cdot c_{2}-b_{2} \\cdot c_{1}}{a_{1} \\cdot b_{2}-a_{2} \\cdot b_{1}}<\/span><\/span><\/p>\n\\Rightarrow x=\\frac{(-5) \\times(-6)-(1) \\times(2)}{(3) \\times(1)-(-2) \\times(-5)}<\/span><\/span><\/p>\n\\Rightarrow x=\\frac{30-2}{3-10}<\/span><\/span><\/p>\n\\Rightarrow x=\\frac{28}{-7}<\/span><\/span><\/p>\n\\Rightarrow x=-4<\/span><\/span><\/p>\n\\mathrm{y}=\\frac{c_{1} \\cdot a_{2}-c_{2} \\cdot a_{1}}{a_{1} \\cdot b_{2}-a_{2} \\cdot b_{1}}<\/span><\/span><\/p>\n\\Rightarrow y=\\frac{(2) \\times(-2)-(-6) \\times(3)}{(3) \\times(1)-(-2) \\times(-5)}<\/span><\/span><\/p>\n\\Rightarrow y=\\frac{(-4)-(-18)}{3-10}<\/span><\/span><\/p>\n\\Rightarrow y=\\frac{14}{-7}<\/span><\/span><\/p>\n\\Rightarrow y=-2<\/span><\/span><\/p>\nTherefore the solution is x = -4 and y = -2.<\/span><\/p>\n[\/et_pb_text][\/et_pb_column][et_pb_column type=”2_5″ module_id=”stickysideR” admin_label=”Column R” _builder_version=”4.10.4″ _module_preset=”default” background_color=”#fdefe0″ custom_padding=”25px|25px|25px|25px|true|true” sticky_position=”top” sticky_offset_top=”-280px” sticky_limit_top=”row” sticky_limit_bottom=”row” sticky_position_tablet=”none” sticky_position_phone=”none” sticky_position_last_edited=”on|desktop” sticky_limit_bottom_tablet=”” sticky_limit_bottom_phone=”” sticky_limit_bottom_last_edited=”on|phone” border_radii=”on|15px|15px|15px|15px” box_shadow_style=”preset3″ global_colors_info=”{}”][et_pb_image src=”https:\/\/eistudymaterial.s3.amazonaws.com\/1080×1080.png” alt=”Free Trial banner” title_text=”Mindspark Free Trial Banner” url=”https:\/\/mindspark.in\/free-trial” align=”center” module_class=”adsimg” _builder_version=”4.11.1″ _module_preset=”default” custom_padding=”||||false|false” border_radii=”on|10px|10px|10px|10px” global_colors_info=”{}” transform_styles__hover_enabled=”on|hover” transform_scale__hover_enabled=”on|hover” transform_translate__hover_enabled=”on|desktop” transform_rotate__hover_enabled=”on|desktop” transform_skew__hover_enabled=”on|desktop” transform_origin__hover_enabled=”on|desktop” transform_scale__hover=”102%|102%”][\/et_pb_image][et_pb_text admin_label=”Explore Other Topics
\n” _builder_version=”4.9.11″ _module_preset=”default” header_font=”|700|||||||” header_font_size=”25px” text_orientation=”center” custom_margin=”0px||0px||true|false” custom_padding=”8px|15px|0px|15px|false|true” locked=”off” global_colors_info=”{}”]<\/p>\n
Explore Other Topics<\/h1>\n
[\/et_pb_text][et_pb_text _builder_version=”4.10.7″ _module_preset=”default” text_line_height=”2.2em” link_font_size=”16px” custom_margin=”||0px||false|false” custom_padding=”10px|15px|10px|28px|true|false” locked=”off” global_colors_info=”{}”]<\/p>\n
\nGeometry<\/a><\/div>\nTrigonometry<\/a><\/div>\nOperations<\/a><\/div>\nNumbers<\/a><\/div>\n<\/div>\n[\/et_pb_text][et_pb_text admin_label=”Related Concepts
\n” _builder_version=”4.9.11″ _module_preset=”default” header_font=”|700|||||||” header_font_size=”25px” text_orientation=”center” custom_margin=”0px||0px||true|false” custom_padding=”8px|15px|0px|15px|false|true” locked=”off” global_colors_info=”{}”]<\/p>\n
Related Concepts<\/h1>\n
[\/et_pb_text][et_pb_text _builder_version=”4.13.1″ _module_preset=”default” text_line_height=”2.2em” link_font_size=”16px” custom_margin=”||0px||false|false” custom_padding=”10px|15px|10px|28px|true|false” locked=”off” global_colors_info=”{}”]<\/p>\n
\nLinear Equation in one variable<\/a><\/div>\nSum and Product of roots<\/a><\/div>\n
<\/span><\/p>\n Example:<\/b><\/p>\n 3x + 2y – 6 = 0\u00a0<\/span><\/p>\n The above equation has two variables i.e, x and y.<\/span><\/p>\n Graphically we can represent this equation by substituting the variables to zero.<\/span><\/p>\n The value of x when y=0 is, we can write the equation as –<\/span><\/p>\n 3x + 2(0) = 6<\/span><\/p>\n \u21d2 x = 2<\/span><\/p>\n We show this solution on the graph as (2,0)<\/span><\/p>\n Therefore the value of y when x = 0 is,<\/span><\/p>\n 3(0) + 2y = 6<\/span><\/p>\n \u21d2 y = 3<\/span><\/p>\n We show this solution on the graph as (0, 3)<\/span><\/p>\n <\/span><\/p>\n <\/span><\/p>\n <\/p>\n Unique Solution –<\/strong> When the lines represented on the coordinate axes by the equations intersect each other, then we say that the pair of equations has a unique solution.<\/p>\n No Solution –<\/strong> When the lines represented on the coordinate axes by the equations do not intersect and are parallel to each other, then we say that the pair of equations has no solution.<\/p>\n Infinite Solutions –<\/strong> When the lines represented on the coordinate axes by the equations are coincident i.e, they both overlap each other ( form the same line) then we say that the pair of equations have infinite solutions.<\/p>\n \n Example:<\/strong> Find the value of x and y by solving the given equations by graphical method. First, we solve each equation separately and plot it. While plotting it on the axes, we find a point of intersection between the two equations. The point of intersection will be the solution.<\/p>\n As we can see in the figure given below, the point of intersection is (3,7).<\/p>\n Therefore the solution is x = 3 and y = 7.<\/p>\n <\/p>\n <\/p>\n Example <\/b>: <\/b>Solve the pair of equations by substitution method:\u00a0<\/span><\/p>\n 5x \u2013 8y = 2\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0– (1)<\/span><\/p>\n \u00a0x + 4y = 3\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 – (2)<\/span><\/p>\n We can\u00a0 pick either of the equations and write one variable in terms of the other, here taking equation (2) first:<\/span><\/p>\n So, we can write\u00a0 x + 4y = 3 as\u00a0 \u00a0 x = 3 – 4y\u00a0 \u00a0 – (3)<\/span><\/p>\n After substituting the value of x in equation (1), we get –<\/span><\/p>\n 5 (3 – 4y) + 8y = 2<\/span><\/p>\n \u21d2 15 – 20y + 8 = 2<\/span><\/p>\n \u21d2 23 – 20y = 2<\/span><\/p>\n \u21d2 – 20y = -21<\/span><\/p>\n \u21d2 y = 21\/20<\/span><\/p>\n Now we substitute the value of y in equation (3),<\/span><\/p>\n x = 3 – 4 (21\/20)<\/span><\/p>\n \u21d2 x = 3 – 21\/5<\/span><\/p>\n \u21d2 x = (15-21)\/5<\/span><\/p>\n \u21d2 x = – 6\/5<\/span><\/p>\n Therefore the solution is x = -6\/5 and y = 21\/20.<\/span><\/p>\n <\/span><\/p>\n <\/b><\/p>\n Example <\/b>:<\/b> Solve the following pair of linear equation by cross multiplication method\u00a0<\/span><\/p>\n 3x – 5y = – 2\u00a0 and y – 2x = 6<\/span><\/p>\n The given equations can be written as:\u00a0<\/span><\/p>\n 3x – 5y +2\u00a0 = 0 \u00a0 \u00a0 – (1)<\/span><\/p>\n -2x + y – 6 = 0 \u00a0 \u00a0 – (2)<\/span><\/p>\n Now by using cross multiplication we get\u00a0<\/span><\/p>\n x=\\frac{b_{1} \\cdot c_{2}-b_{2} \\cdot c_{1}}{a_{1} \\cdot b_{2}-a_{2} \\cdot b_{1}}<\/span><\/span><\/p>\n \\Rightarrow x=\\frac{(-5) \\times(-6)-(1) \\times(2)}{(3) \\times(1)-(-2) \\times(-5)}<\/span><\/span><\/p>\n \\Rightarrow x=\\frac{30-2}{3-10}<\/span><\/span><\/p>\n \\Rightarrow x=\\frac{28}{-7}<\/span><\/span><\/p>\n \\Rightarrow x=-4<\/span><\/span><\/p>\n \\mathrm{y}=\\frac{c_{1} \\cdot a_{2}-c_{2} \\cdot a_{1}}{a_{1} \\cdot b_{2}-a_{2} \\cdot b_{1}}<\/span><\/span><\/p>\n \\Rightarrow y=\\frac{(2) \\times(-2)-(-6) \\times(3)}{(3) \\times(1)-(-2) \\times(-5)}<\/span><\/span><\/p>\n \\Rightarrow y=\\frac{(-4)-(-18)}{3-10}<\/span><\/span><\/p>\n \\Rightarrow y=\\frac{14}{-7}<\/span><\/span><\/p>\n \\Rightarrow y=-2<\/span><\/span><\/p>\n Therefore the solution is x = -4 and y = -2.<\/span><\/p>\n [\/et_pb_text][\/et_pb_column][et_pb_column type=”2_5″ module_id=”stickysideR” admin_label=”Column R” _builder_version=”4.10.4″ _module_preset=”default” background_color=”#fdefe0″ custom_padding=”25px|25px|25px|25px|true|true” sticky_position=”top” sticky_offset_top=”-280px” sticky_limit_top=”row” sticky_limit_bottom=”row” sticky_position_tablet=”none” sticky_position_phone=”none” sticky_position_last_edited=”on|desktop” sticky_limit_bottom_tablet=”” sticky_limit_bottom_phone=”” sticky_limit_bottom_last_edited=”on|phone” border_radii=”on|15px|15px|15px|15px” box_shadow_style=”preset3″ global_colors_info=”{}”][et_pb_image src=”https:\/\/eistudymaterial.s3.amazonaws.com\/1080×1080.png” alt=”Free Trial banner” title_text=”Mindspark Free Trial Banner” url=”https:\/\/mindspark.in\/free-trial” align=”center” module_class=”adsimg” _builder_version=”4.11.1″ _module_preset=”default” custom_padding=”||||false|false” border_radii=”on|10px|10px|10px|10px” global_colors_info=”{}” transform_styles__hover_enabled=”on|hover” transform_scale__hover_enabled=”on|hover” transform_translate__hover_enabled=”on|desktop” transform_rotate__hover_enabled=”on|desktop” transform_skew__hover_enabled=”on|desktop” transform_origin__hover_enabled=”on|desktop” transform_scale__hover=”102%|102%”][\/et_pb_image][et_pb_text admin_label=”Explore Other Topics [\/et_pb_text][et_pb_text _builder_version=”4.10.7″ _module_preset=”default” text_line_height=”2.2em” link_font_size=”16px” custom_margin=”||0px||false|false” custom_padding=”10px|15px|10px|28px|true|false” locked=”off” global_colors_info=”{}”]<\/p>\n [\/et_pb_text][et_pb_text admin_label=”Related Concepts [\/et_pb_text][et_pb_text _builder_version=”4.13.1″ _module_preset=”default” text_line_height=”2.2em” link_font_size=”16px” custom_margin=”||0px||false|false” custom_padding=”10px|15px|10px|28px|true|false” locked=”off” global_colors_info=”{}”]<\/p>\n<\/span><\/p>\n<\/span><\/h2>\n
Types of Solution of Pair of Linear Equation with Two Variables-<\/strong><\/h2>\n
<\/span><\/p>\nMethods To Solve Pair Of Linear Equation In Two Variables-<\/strong><\/h2>\n
1. Graphical Method :<\/h3>\n
\n
x – y + 4 = 0 – (1) and x + y – 10 = 0 – (2)<\/p>\n<\/p>\n2. Substitution Method :<\/h3>\n
\n
3. Cross Multiplication Method :<\/span><\/h3>\n
\n
<\/span><\/p>\n\n
<\/span><\/p>\n\n
<\/span><\/p>\n
\n” _builder_version=”4.9.11″ _module_preset=”default” header_font=”|700|||||||” header_font_size=”25px” text_orientation=”center” custom_margin=”0px||0px||true|false” custom_padding=”8px|15px|0px|15px|false|true” locked=”off” global_colors_info=”{}”]<\/p>\nExplore Other Topics<\/h1>\n
\n” _builder_version=”4.9.11″ _module_preset=”default” header_font=”|700|||||||” header_font_size=”25px” text_orientation=”center” custom_margin=”0px||0px||true|false” custom_padding=”8px|15px|0px|15px|false|true” locked=”off” global_colors_info=”{}”]<\/p>\nRelated Concepts<\/h1>\n